POI2015 解题报告

由于博主没有BZOJ权限号，是在洛咕做的题~

完成了13题（虽然有一半难题都是看题解的QAQ）剩下的题咕咕咕~~

Luogu3585 [POI2015]PIE

Solution

模拟，按顺序搜索，把搜索到的需要印却没有印的点和印章的第一个点重合，并印上。

另外，纸上需要印的点和印章上沾墨水的点用数组储存，能加快很多

Code

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #include<vector>
 5 #define R register
 6 using namespace std;
 7 typedef pair<int, int> P;
 8 
 9 const int N = 1e3 + 5;
10 
11 int n, m, a, b;
12 int stx, sty;
13 char s[N], in[N][N], mp[N][N];
14 
15 vector<P> need, offer;
16 
17 int jud(int x, int y) {
18     if (x <= 0 || y <= 0 || x > n || y > m)
19         return 0;
20     return 1;
21 }
22 
23 #define X first
24 #define Y second
25 int col(int x, int y) {
26     for (R int i = 0, up = offer.size(); i < up; ++i) {
27         int onx = x + offer[i].X - stx, ony = y + offer[i].Y - sty;
28         if (!jud(onx, ony)) return 0;
29         if (mp[onx][ony] == '.') return 0;
30         mp[onx][ony] = '.';
31     }
32     return 1;
33 }
34 
35 int work() {
36     stx = sty = 0;
37     offer.clear(); need.clear();
38     for (R int i = 1; i <= n; ++i) scanf("%s", mp[i] + 1);
39     for (R int i = 1; i <= a; ++i) scanf("%s", in[i] + 1);
40     for (R int i = 1; i <= n; ++i)
41         for (R int j = 1; j <= m; ++j) if (mp[i][j] == 'x')
42             need.push_back(P(i, j));
43     for (R int i = 1; i <= a; ++i)
44         for (R int j = 1; j <= b; ++j) if (in[i][j] == 'x') {
45             offer.push_back(P(i, j));
46             if (!stx) stx = i, sty = j;
47         }
48     
49     for (int i = 0, up = need.size(); i < up; ++i)
50         if (mp[need[i].X][need[i].Y] == 'x')
51             if (!col(need[i].X, need[i].Y)) return 0;
52     return 1;
53 }
54 #undef X
55 #undef Y
56 
57 int main()
58 {
59     int Q; scanf("%d", &Q);
60     for (; Q; Q--) {
61         scanf("%d%d%d%d", &n, &m, &a, &b);
62         if (work()) puts("TAK");
63         else puts("NIE");
64     }
65 }

Luogu3585 PIE

Luogu3594[POI2015]WIL-Wilcze doły

Solution

单调队列，将长度为 $d$ 的最大字段和加入队列，并且队列内字段和单调递减

开个双指针 $i, j$ 表示要选择的最长的连续区间的两端。

随着 $i$ 增加，把新的长度为$d$ 的子段和加入队列。

然后逐渐右移指针$j$，直到找到第一个$<=p$的区间。随着$j$增加，把队列内超出范围的子段和弹出

Code

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #define rd read()
 5 #define ll long long
 6 #define R register
 7 using namespace std;
 8 
 9 const int N = 2e6 + 5;
10 
11 int n, p, d;
12 ll sum[N], a[N];
13 ll q[N];
14 
15 ll read() {
16     ll X = 0, p = 1; char c = getchar();
17     for (; c > '9' || c < '0'; c = getchar())
18         if (c == '-') p = -1;
19     for (; c >= '0' && c <= '9'; c = getchar())
20         X = X * 10 + c - '0';
21     return X * p;
22 }
23 
24 int main()
25 {
26     n = rd; p = rd; d = rd;
27     for (R int i = 1; i <= n; ++i)
28         a[i] = rd, sum[i] = sum[i - 1] + a[i];
29     int l = 1, r = 0, ans = d;
30     for (int i = d, j = 0; i <= n; ++i) {
31         ll tmp = sum[i] - sum[i - d];
32         while (l <= r && sum[q[r]] - sum[q[r] - d] <= tmp) r--;
33         q[++r] = i;
34         tmp = sum[q[l]] - sum[q[l] - d];
35         while (sum[i] - sum[j] - tmp > p) {
36             j++;
37             while (l <= r && q[l] - d < j) l++;
38             tmp = sum[q[l]] - sum[q[l] - d];
39         }
40         ans = max(ans, i - j);
41     }
42     printf("%d
", ans);
43 }

Luogu3594 WIL-Wilcze doły

Luogu3586[POI2015]LOG

Solution

树状数组

先考虑怎样判断是否符合条件，数列中 $>=s$的个数为$cnt$，若剩余的$<s$的数的和 $>= (c-cnt)*s$ 即可满足条件

这样我们就需要知道数列中有多少个数$>=s$，以及$<s$的数的和，可以用两个树状数组维护.

最后一个点开LL

Code

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #define rd read()
 5 #define ll long long
 6 using namespace std;
 7 
 8 const int N = 2e6 + 5;
 9 
10 ll a[N], n, m, tot;
11 ll cnt[N], ls[N], sum[N];
12 
13 struct node {
14     int typ, x;
15     ll y;
16 }pro[N];
17 
18 ll read() {
19     ll X = 0, p = 1; char c = getchar();
20     for (; c > '9' || c < '0'; c = getchar())
21         if (c == '-') p = -1;
22     for (; c >= '0' && c <= '9'; c = getchar())
23         X = X * 10 + c - '0';
24     return X * p;
25 }
26 
27 int lowbit(int x) {
28     return x & -x;
29 }
30 
31 template <typename T>
32 void add(int x, ll d, T *s) {
33     for (; x <= tot; x += lowbit(x))
34         s[x] += d;
35 }
36 
37 template <typename T>
38 T query(int x, T *s) {
39     T re = 0;
40     for (; x; x -= lowbit(x))
41         re += s[x];
42     return re;
43 }
44 
45 int fd(ll x) {
46     return lower_bound(ls + 1, ls + 1 + tot, x) - ls;
47 }
48 
49 int main()
50 {
51     n = rd; m = rd;
52     tot = 1;
53     for (int i = 1; i <= m; ++i) {
54         char ch = getchar();
55         while (ch > 'Z' || ch < 'A') ch = getchar();
56         if (ch == 'U') {
57             pro[i].typ = 1; pro[i].x = rd; pro[i].y = rd;
58             ls[++tot] = pro[i].y;
59         }
60         else {
61             pro[i].typ = 2; pro[i].x = rd; pro[i].y = rd;
62         }
63     }
64     sort(ls + 1, ls + 1 + tot);
65     tot = unique(ls + 1, ls + 1 + tot) - ls - 1;
66     for (int i = 1; i <= n; ++i)
67         add(1, 1, cnt);
68     for (int i = 1; i <= m; ++i) {
69         if (pro[i].typ == 1) {
70             int ch = fd(a[pro[i].x]);
71             add(ch, -1, cnt);
72             add(ch, -ls[ch], sum);
73             ch = fd(pro[i].y);
74             add(ch, 1, cnt);
75             add(ch, ls[ch], sum);
76             a[pro[i].x] = pro[i].y;
77         }
78         else {
79             int ch = fd(pro[i].y), num;
80             num = n - query(ch - 1, cnt);
81             ll tmp = (pro[i].x - num) * pro[i].y;
82             if (query(ch - 1, sum) >= tmp)
83                 puts("TAK");
84             else puts("NIE");
85         }
86     }
87 }

Luogu3586 LOG

Luogu3584[POI2015]LAS

Solution

环形DP

$S$ 表示第$i$个人以及和他相邻的两个人吃哪边的食物，例如$S$的二进制上有4，就表示第$i-1$个人吃右边的食物，反之，则吃左边的食物

设置状态$f[i][S]$ 表示第$i$ 个人，他相邻的吃食物的情况为$S$，能否符合要求。

由于环形最后一个人会影响第一个人，则先枚举第一个人，到最后一个人判断是否存在与第一个人状态相符的情况符合要求。

Code

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 #define rd read()
 5 #define db double
 6 using namespace std;
 7 
 8 const int N = 1e6 + 5;
 9 
10 int n, c[N], f[N][8], ans[N], path[N][8];
11 
12 int read() {
13     int X = 0, p = 1; char c = getchar();
14     for (; c > '9' || c < '0'; c = getchar())
15         if (c == '-') p = -1;
16     for (; c >= '0' && c <= '9'; c = getchar())
17         X = X * 10 + c - '0';
18     return X * p;
19 }
20 
21 int ch(int x) {
22     return (x + n) % n;
23 }
24 
25 int jud(int x, int S) {
26     int le, re, me;
27     le = (S >> 2) & 1; re = S & 1; me = (S >> 1) & 1;
28     db now = c[ch(x + me)], nxt;
29     if (!me && le) now /= 2;
30     if (me && !re) now /= 2;
31     nxt = c[ch(x + (!me))];
32     if (me && le) nxt /= 2;
33     if (!me && !re) nxt /= 2;
34     if (nxt > now) return 0;
35     else return 1;
36 }
37 
38 
39 int work(int S) {
40     memset(f, 0, sizeof (f));
41     f[0][S] = 1;
42     for (int i = 1; i < n; ++i) 
43         for (int now = 0; now < 8; ++now) if (jud(i, now)) 
44             for (int pre = 0; pre <= 4; pre += 4) if (f[i - 1][pre + (now >> 1)])
45                 f[i][now] = 1, path[i][now] = pre + (now >> 1);
46     if (!f[n - 1][S >> 1] && !f[n - 1][(S >> 1) + 4])
47         return 0;
48     if (f[n - 1][S >> 1]) {
49         for (int now = S >> 1, i = n - 1; ~i; now = path[i][now], --i)
50             ans[i] = (now & 2) >> 1;
51         for (int i = 0; i < n; ++i)
52             printf("%d ", (ans[i] + i) % n + 1);
53     }
54     else {
55         for (int now = (S >> 1) + 4, i = n - 1; ~i; now = path[i][now], --i)
56             ans[i] = (now & 2) >> 1;
57         for (int i = 0; i < n; ++i)
58             printf("%d ", (ans[i] + i) % n + 1);
59     }
60     return 1;
61 }
62 
63 int main()
64 {
65     n = rd;
66     for (int i = 0; i < n; ++i)
67         c[i] = rd;
68     for (int i = 0; i < 8; ++i) if(jud(0, i))
69         if (work(i)) return 0;
70     puts("NIE");
71 }

Luogu3584 LAS

Luogu3588[POI2015]PUS

Solution

线段树优化建树+差分约束

我们最初的想法应该是在区间$[L,R]$内选中的数向未被选中的数连一条长度为$1$的边，一次操作便有$N^2$条边，这样肯定会MLE+TLE

于是我们又想到另外建一个虚点，被选中的数向虚点建一条长度为$1$ 的边，虚点再向未被选中的数连长度为0的边，这样一次操作便有$N$条边，仍会MLE+TLE

于是我们用线段树优化建图，所有操作中区间约有$k+1$段，所以虚点向区间连边，被选中的点再向虚点连边。就解决了这个问题。

最后再差分约束一下。

Code

  1 #include<cstdio>
  2 #include<cstring>
  3 #include<algorithm>
  4 #include<queue>
  5 #define rd read()
  6 using namespace std;
  7 
  8 const int N = 4e5 + 5;
  9 
 10 int head[N], tot;
 11 int n, dis[N], m, p, vis[N], r[N];
 12 int pre[N];
 13 
 14 struct edge {
 15     int nxt, to, w;
 16 }e[N << 2];
 17 
 18 queue<int> q;
 19 
 20 int read() {
 21     int X = 0, p = 1; char c = getchar();
 22     for (; c > '9' || c < '0'; c = getchar())
 23         if (c == '-') p = -1;
 24     for (; c >= '0' && c <= '9'; c = getchar())
 25         X = X * 10 + c - '0';
 26     return X * p;
 27 }
 28 
 29 void add(int u, int v, int w) {
 30     e[++tot].to = v;
 31     e[tot].nxt = head[u];
 32     e[tot].w = w;
 33     r[v]++;
 34     head[u] = tot;
 35 }
 36 
 37 namespace SegT {
 38     int lc[N], rc[N], cnt, root;
 39 #define mid ((l + r) >> 1)
 40     
 41     void build(int &x, int l, int r) {
 42         if (l == r) {
 43             x = l; return;
 44         }
 45         x = ++cnt;
 46         build(lc[x], l, mid);
 47         build(rc[x], mid + 1, r);
 48         add(lc[x], x, 0);
 49         add(rc[x], x, 0);
 50     }
 51 
 52     void update(int L, int R, int c, int l, int r, int x) {
 53         if (L > R) return;
 54         if (L <= l && r <= R) {
 55             add(x, c, 0); return;
 56         }
 57         if (mid >= L)
 58             update(L, R, c, l, mid, lc[x]);
 59         if (mid < R)
 60             update(L, R, c, mid + 1, r, rc[x]);
 61     }
 62 }using namespace SegT;
 63 
 64 void cmax(int &A, int B) {
 65     if (A < B)
 66         A = B;
 67 }
 68 
 69 void Topsort() {
 70     for (int i = 1; i <= cnt; ++i) {
 71         if (!dis[i]) dis[i] = 1;
 72         if (!r[i]) q.push(i);
 73     }
 74     for (int u; !q.empty();) {
 75         u = q.front(); q.pop();
 76         vis[u] = 1;
 77         for (int i = head[u]; i; i = e[i].nxt) {
 78             int nt = e[i].to;
 79             cmax(dis[nt], dis[u] + e[i].w);
 80             if (!(--r[nt])) q.push(nt);
 81         }
 82     }
 83 }
 84 
 85 int main()
 86 {
 87     cnt = n = rd; p = rd; m = rd;
 88     for (int i = 1; i <= p; ++i) {
 89         int pos = rd, x = rd;
 90         pre[pos] = dis[pos] = x;
 91     }
 92     build(root, 1, n);
 93     for (; m; m--) {
 94         int l = rd, r = rd, num = rd;
 95         int last = l, now;
 96         ++cnt;
 97         for (; num; --num) {
 98             add(cnt, now = rd, 1);
 99             update(last, now - 1, cnt, 1, n, root);
100             last = now + 1;
101         }
102         update(now + 1, r, cnt, 1, n, root);
103     }
104     Topsort();
105     for (int i = 1; i <= n; ++i) 
106         if (!vis[i] || dis[i] > 1e9 || (dis[i] > pre[i] && pre[i]))
107             return puts("NIE"), 0;
108     puts("TAK");
109     for (int i = 1; i <= n; ++i)
110         printf("%d ", dis[i]);
111 }

Luogu3588 PUS

Luogu3596[POI2015]MOD

Solution

树形DP

确实有难度啊QAQ

要求出每个子树的直径，以及删去子树后剩下的那棵树的直径。

要使合并后的树直径最小，需要把两棵树的直径的中点连起来，设两个直径分别为 $f, g$最后得到的直径为 $max{f, g, (f+1)/2+(g+1)/2+1}$

要使合并后的树直径最大，则把直径两端给连起来，为$f+g+1$

状态转移不好讲，就讲变量的意义，具体看代码里的两个$dp$

$f[i]$ 表示以$i$为根的子树的直径

$w[i][0]$ 表示以 $i$的子节点为根的子树的直径中最大的直径

$w[i][1]$ 表示以 $i$的子节点为根的子树的直径中第二大的直径

$d[i][0]$ 表示从 $i$ 出发往下的最长链的长度

$d[i][1]$ 和 $d[i][2]$ 依次类推

$line[i]$ 表示从$i$开始，到除$i$外的子节点所能得到的最长链

$g[i]$ 表示删去以$i$ 为节点的子树后得到的树的直径

最后要得到方案：

直径最长则 $bfs$ 求出两条直径的端点

直径最短，则先求出两条直径的端点，然后往上跳，找到直径的中点

总复杂度$O(N)$

Code

  1 #include<cstdio>
  2 #include<cstring>
  3 #include<algorithm>
  4 #include<queue>
  5 #define rd read()
  6 const int N = 5e5 + 5;
  7 using namespace std;
  8 
  9 int f[N], g[N], d[N][3], w[N][2], line[N], n, fa[N];
 10 int head[N], tot;
 11 int ansmax, ansmin = N, maxx, maxy, minx, miny;
 12 int diax1, diay1, diax2, diay2;
 13 int dep[N], vis[N];
 14 
 15 queue<int> q;
 16 
 17 struct edge {
 18     int nxt, to;
 19 }e[N << 1];
 20 
 21 int read() {
 22     int X = 0, p = 1; char c = getchar();
 23     for (; c > '9' || c < '0'; c = getchar())
 24         if (c == '-') p = -1;
 25     for (; c >= '0' && c <= '9'; c = getchar())
 26         X = X * 10 + c - '0';
 27     return X * p;
 28 }
 29 
 30 void add(int u, int v) {
 31     e[++tot].to = v;
 32     e[tot].nxt = head[u];
 33     head[u] = tot;
 34 }
 35 
 36 void cmax(int &A, int B) {
 37     if (A < B)
 38         A = B;
 39 }
 40 
 41 void cmin(int &A, int B) {
 42     if (A > B)
 43         A = B;
 44 }
 45 
 46 void dp1(int u) {
 47     for (int i = head[u]; i; i = e[i].nxt) {
 48         int nt = e[i].to;
 49         if (nt == fa[u]) continue;
 50         dep[nt] = dep[u] + 1;
 51         fa[nt] = u;
 52         dp1(nt);
 53         cmax(f[u], f[nt]);
 54         int tmp = d[nt][0] + 1;
 55         if (tmp > d[u][0])
 56             d[u][2] = d[u][1], d[u][1] = d[u][0], d[u][0] = tmp;
 57         else if (tmp > d[u][1])
 58             d[u][2] = d[u][1], d[u][1] = tmp;
 59         else if (tmp > d[u][2])
 60             d[u][2] = tmp;
 61         tmp = f[nt];
 62         if (tmp > w[u][0])
 63             w[u][1] = w[u][0], w[u][0] = tmp;
 64         else if (tmp > w[u][1])
 65             w[u][1] = tmp;
 66     }
 67     cmax(f[u], d[u][0] + d[u][1]);
 68 }
 69 
 70 void dp2(int u) {
 71     if (u != 1) {
 72         if (ansmax < g[u] + f[u] + 1) {
 73             ansmax = g[u] + f[u] + 1;
 74             maxx = fa[u]; maxy = u;
 75         }
 76         int res = max(g[u], f[u]);
 77         cmax(res, (g[u] + 1) / 2 + (f[u] + 1) / 2 + 1);
 78         if (ansmin > res) {
 79             ansmin = res;
 80             minx = fa[u]; miny = u;
 81         }
 82     }
 83     for (int i = head[u]; i; i = e[i].nxt) {
 84         int nt = e[i].to;
 85         if (nt == fa[u]) continue;
 86         cmax(line[nt], line[u] + 1);
 87         cmax(g[nt], g[u]);
 88         int tmp = d[nt][0] + 1;
 89         if (tmp == d[u][0]) {
 90             cmax(g[nt], d[u][1] + d[u][2]);
 91             cmax(g[nt], line[u] + d[u][1]);
 92             cmax(line[nt], d[u][1] + 1);
 93         }
 94         else if (tmp == d[u][1]) {
 95             cmax(g[nt], d[u][0] + d[u][2]);
 96             cmax(g[nt], line[u] + d[u][0]);
 97             cmax(line[nt], d[u][0] + 1);
 98         }
 99         else {
100             cmax(g[nt], d[u][0] + d[u][1]);
101             cmax(g[nt], line[u] + d[u][0]);
102             cmax(line[nt], d[u][0] + 1);
103         }
104         tmp = f[nt];
105         if (tmp == w[u][0]) 
106             cmax(g[nt], w[u][1]);
107         else cmax(g[nt], w[u][0]);
108         dp2(nt);
109     }
110 }
111 
112 void outputmax() {
113     memset(vis, 0, sizeof(vis));
114     printf("%d %d %d", ansmax, maxx, maxy);
115     q.push(maxx);
116     vis[maxx] = 1;
117     for (int u; !q.empty();) {
118         u = q.front(); q.pop();
119         maxx = u;
120         for (int i = head[u]; i; i = e[i].nxt) {
121             int nt = e[i].to;
122             if (nt == maxx || nt == maxy) continue;
123             if (vis[nt]) continue;
124             q.push(nt);
125             vis[nt] = 1;
126         }
127     }
128     q.push(maxy);
129     vis[maxy] = 1;
130     for (int u; !q.empty();) {
131         u = q.front(); q.pop();
132         maxy = u;
133         for (int i = head[u]; i; i = e[i].nxt) {
134             int nt = e[i].to;
135             if (vis[nt]) continue;
136             q.push(nt);
137             vis[nt] = 1;
138         }
139     }
140     printf(" %d %d
", maxx, maxy);
141 }
142 
143 int bfs(int S) {
144     memset(vis, 0, sizeof(vis));
145     q.push(S);
146     vis[S] = 1;
147     int re;
148     for (int u; !q.empty();) {
149         u = q.front(); q.pop();
150         re = u;
151         for (int i = head[u]; i; i = e[i].nxt) {
152             int nt = e[i].to;
153             if ((u == minx && nt == miny) || (u == miny && nt == minx))
154                 continue;
155             if (vis[nt]) continue;
156             q.push(nt);
157             vis[nt] = 1;
158         }
159     }
160     return re;
161 }
162 
163 int solve(int x, int y, int len) {
164     int rest = len;
165     if (dep[x] < dep[y]) swap(x, y);
166     while (rest != (len + 1) / 2)
167         x = fa[x], rest --;
168     return x;
169 }
170 
171 int main()
172 {
173     n = rd; 
174     for (int i = 1; i < n; ++i) {
175         int u = rd, v = rd;
176         add(u, v); add(v, u);
177     }
178     dp1(1); dp2(1);
179     printf("%d %d %d", ansmin, minx, miny);
180     diax1 = bfs(minx); diay1 = bfs(diax1);
181     diax2 = bfs(miny); diay2 = bfs(diax2);
182     printf(" %d %d
", solve(diax1, diay1, g[miny]), solve(diax2, diay2, f[miny]));
183     outputmax();
184 }

Luogu3596 MOD

Luogu3592 [POI2015]MYJ

区间DP

题解传送门

Luogu3590[POI2015]TRZ

Solution

若子串中只存在一种字符 : $O(N)$扫一遍即可得到答案

其他情况：设sum[i][k] 为前$i$ 个字符中，字符$k$的个数

若一个子串$[j + 1, i]$满足题设条件，则

　　$sum[i][1] - sum[j][1] != sum[i][2] - sum[j][2]$

　　$sum[i][2] - sum[j][2] != sum[i][3] - sum[j][3]$

　　$sum[i][3] - sum[j][3] != sum[i][1] - sum[j][1]$

移项得到：

　　$sum[i][1] - sum[i][2] != sum[j][1] - sum[j][2]$

　　$sum[i][2] - sum[i][3] != sum[j][2] - sum[j][3]$

　　$sum[i][3] - sum[i][1] != sum[j][3] - sum[j][1]$

设 $x = sum[i][1] - sum[i][2], y = sum[i][2] - sum[i][3], z = sum[i][3] - sum[i][1]$

则要满足 $x_i != x_j, y_i != y_j, z_i != z_j$

有三维，直接求解会很麻烦。

用线段树或者树状数组维护（我不会用树状数组QAQ）

首先对$x$进行排序，相同的$x$一起查询更新，这样就少了一维

然后以 $y$ 为线段树的下标，

节点内存储区间的最小的标号 $i$ ，次小标号，最小标号的 $z$ 值，最大标号，次大标号，最大标号的$z$值

最后吸氧才过的呜呜呜，常数太大了

Code

  1 // luogu-judger-enable-o2
  2 #include<cstdio>
  3 #include<cstring>
  4 #include<algorithm>
  5 #define rd read()
  6 #define N 1000005
  7 #define inf 1e8
  8 #define ri register
  9 using namespace std;
 10 
 11 int n, sum[4], ans, ls[N], tot;
 12 char op[N];
 13 
 14 struct P {
 15     int id, x, y, z;
 16 }a[N];
 17 
 18 inline int read() {
 19     int X = 0, p = 1; char c = getchar();
 20     for (; c > '9' || c < '0'; c = getchar())
 21         if (c == '-') p = -1;
 22     for (; c >= '0' && c <= '9'; c = getchar())
 23         X = X * 10 + c - '0';
 24     return X * p;
 25 }
 26 
 27 inline int cmp(const P &A, const P &B) {
 28     return A.x < B.x;
 29 }
 30 
 31 inline void cmax(int &A, int B) {
 32     if (A < B)
 33          A = B;
 34 }
 35 
 36 namespace SegT {
 37 #define mid ((l + r) >> 1)
 38 #define lson x << 1
 39 #define rson x << 1 | 1
 40     struct node {
 41         int maxz, minz, maxn[2], minn[2];
 42         node () {
 43             maxn[0] = maxn[1] = -inf;
 44             minn[0] = minn[1] = inf;
 45             maxz = minz = inf;
 46         }
 47     }pt[N << 2];
 48     
 49     inline void up(node&x, int val, int z) {
 50         if (val > x.maxn[0] && z != x.maxz) {
 51             x.maxn[1] = x.maxn[0];
 52             x.maxn[0] = val;
 53             x.maxz = z;
 54         }
 55         else if (val > x.maxn[0]&& z == x.maxz) {
 56             x.maxn[0] = val;
 57             x.maxz = z;
 58         }
 59         else if (val > x.maxn[1] && z != x.maxz) {
 60             x.maxn[1] = val;
 61         }
 62     }
 63 
 64     inline void down(node &x, int val, int z) {
 65         if (val < x.minn[0] && z != x.minz) {
 66             x.minn[1] = x.minn[0];
 67             x.minn[0] = val;
 68             x.minz = z;
 69         }
 70         else if (val < x.minn[0] && z == x.minz) {
 71             x.minn[0] = val;
 72             x.minz = z;
 73         }
 74         else if (val < x.minn[1] && z != x.minz) {
 75             x.minn[1] = val;
 76         }
 77     }
 78 
 79     node merge(node l, node r) {
 80         node res = l;
 81         up(res, r.maxn[0], r.maxz);
 82         up(res, r.maxn[1], inf);
 83         down(res, r.minn[0], r.minz);
 84         down(res, r.minn[1], inf);
 85         return res;
 86     }
 87 
 88     inline void pushup(int x) {
 89         pt[x] = merge(pt[lson], pt[rson]);
 90     }
 91 
 92     void modify(int pos, int val, int z, int l, int r, int x) {
 93         if (l == r) {
 94             up(pt[x], val, z); down(pt[x], val, z);
 95             return;
 96         }
 97         if (pos <= mid)
 98             modify(pos, val, z, l, mid, lson);
 99         else modify(pos, val, z, mid + 1, r, rson);
100         pushup(x);
101     }
102 
103     node query(int L, int R, int l, int r, int x) {
104         if (L <= l && r <= R)
105             return pt[x];
106         node res, ltmp, rtmp;
107         if (L > R) return res;
108         if (mid >= L)
109             ltmp = query(L, R, l, mid, lson);
110         if (mid < R)
111             rtmp = query(L, R, mid + 1, r, rson);
112         res = merge(ltmp, rtmp);
113         return res;
114     }
115 }using namespace SegT;
116 
117 int main()
118 {
119     n = rd; tot = 1;
120     scanf("%s", op + 1);
121     for (ri int i = 1, last = 0; i <= n; ++i) {
122         int typ;
123         if (op[i] == 'B') typ = 1;
124         else if (op[i] == 'C') typ = 2;
125         else typ = 3;
126         sum[typ]++;
127         a[i].id = i;
128         a[i].x = sum[1] - sum[2];
129         a[i].y = sum[2] - sum[3];
130         a[i].z = sum[3] - sum[1];
131         ls[++tot] = a[i].y;
132         if (typ == last) sum[0]++;
133         else sum[0] = 1;
134         cmax(ans, sum[0]);
135     }
136     a[0].x = a[0].y = a[0].z = a[0].id = 0;
137     sort(a, a + 1 + n, cmp);
138     sort(ls + 1, ls + 1 + tot);
139     tot = unique(ls + 1, ls + 1 + tot) - ls - 1;
140     for (ri int i = 0; i <= n; ++i)
141         a[i].y = lower_bound(ls + 1, ls + 1 + tot, a[i].y) - ls;
142     a[n + 1].x = inf;
143     for (ri int i = 0, j = 0; i <= n; i = j + 1) {
144         for (j = i; j <= n && a[j].x == a[j + 1].x; ++j) {
145             node ltmp = query(1, a[j].y - 1, 1, tot, 1);
146             node rtmp = query(a[j].y + 1, tot, 1, tot, 1);
147             node tmp = merge(ltmp, rtmp);
148             if (tmp.maxz == a[j].z)
149                 cmax(ans, tmp.maxn[1] - a[j].id);
150             else cmax(ans, tmp.maxn[0] - a[j].id);
151             if (tmp.minz == a[j].z)
152                 cmax(ans, a[j].id - tmp.minn[1]);
153             else cmax(ans, a[j].id - tmp.minn[0]);
154         }
155 
156         node ltmp = query(1, a[j].y - 1, 1, tot, 1);
157         node rtmp = query(a[j].y + 1, tot, 1, tot, 1);
158         node tmp = merge(ltmp, rtmp);
159         if (tmp.maxz == a[j].z)
160             cmax(ans, tmp.maxn[1] - a[j].id);
161         else cmax(ans, tmp.maxn[0] - a[j].id);
162         if (tmp.minz == a[j].z)
163             cmax(ans, a[j].id - tmp.minn[1]);
164         else cmax(ans, a[j].id - tmp.minn[0]);
165 
166         for (j = i; j <= n && a[j].x == a[j + 1].x; ++j) {
167             modify(a[j].y, a[j].id, a[j].z, 1, tot, 1);
168         }
169         modify(a[j].y, a[j].id, a[j].z, 1, tot, 1);
170     }
171     printf("%d
", ans);
172 }

Luogu3590 TRZ

Luogu3591[POI2015]ODW

传送门~~

Luogu3597[POI2015]WYC

传送门~~

Luogu3587[POI2015]POD

不出意外又去看题解了，是比较套路和有思维的题目

大佬写的题解简单易懂： 传送门 (https://www.cnblogs.com/five20/p/9581552.html) byfive20

环形前缀和，用hash记录每种颜色出现的次数，在某种颜色的最后一个位置删去该颜色的全部贡献。

两个hash值相同的位置就是可以分割的点。因为要$r-l$ 尽可能接近mid ，能用单调队列来维护。

Code

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<cstring>
 4 #define rd read()
 5 #define ll long long
 6 #define R register
 7 using namespace std;
 8 
 9 const int N = 1000005, base1 = 999979, base2 = 999983, mod1 = 1e9 + 7, mod2 = 1e9 + 9;
10 
11 ll po1[N], po2[N], sum1, sum2;
12 int last[N], cnt[N], n, k, a[N];
13 
14 struct node {
15     int id;
16     ll s1, s2;
17 }b[N];
18 
19 inline int read() {
20     int X = 0, p = 1; char c = getchar();
21     for (; c > '9' || c < '0'; c = getchar())
22         if (c == '-') p = -1;
23     for (; c >= '0' && c <= '9'; c = getchar())
24         X = X * 10 + c - '0';
25     return X * p;
26 }
27 
28 int cmp(const node &A, const node &B) {
29     if (A.s1 != B.s1) return A.s1 < B.s1;
30     if (A.s2 != B.s2) return A.s2 < B.s2;
31     return A.id < B.id;
32 }
33 
34 int jud(int x, int y) {
35     if (b[x].s1 != b[y].s1) return 0;
36     if (b[x].s2 != b[y].s2) return 0;
37     return 1;
38 }
39 
40 void cmin(int &A, int B) {
41     if (A > B) A = B;
42 }
43 
44 int Abs(int A) {
45     return A > 0 ? A : -A;
46 }
47 
48 int main()
49 {
50     n = rd; k = rd;
51     for (R int i = 1; i <= n; ++i)
52         a[i] = rd;
53     po1[0] = po2[0] = 1;
54     for (R int i = 1; i <= k; ++i)
55         po1[i] = po1[i - 1] * base1 % mod1,
56         po2[i] = po2[i - 1] * base2 % mod2;
57     for (R int i = 1; i <= n; ++i) 
58         cnt[a[i]]++, last[a[i]] = i;
59     for (R int i = 1; i <= n; ++i) {
60         (sum1 += po1[a[i]]) %= mod1;
61         (sum2 += po2[a[i]]) %= mod2;
62         if (i == last[a[i]]) 
63             sum1 = (sum1 - po1[a[i]] * cnt[a[i]] % mod1) % mod1,
64             sum1 = (sum1 + mod1) % mod1,
65             sum2 = (sum2 - po2[a[i]] * cnt[a[i]] % mod2) % mod2,
66             sum2 = (sum2 + mod2) % mod2;
67         b[i].id = i,
68         b[i].s1 = sum1,
69         b[i].s2 = sum2;
70     }
71     sort(b + 1, b + 1 + n, cmp);
72     ll ans1 = 0; int ans2 = n;
73     int mid = (n + 1) >> 1;
74     for (int i = 1, j = 1; i <= n; i = j) {
75         while (j <= n && jud(i, j)) j++;
76         ans1 += 1LL * (j - i) * (j - i - 1) / 2;
77         for (int l = i, r = i; r < j; ++r) {
78             while (l < r && b[r].id - b[l].id >= mid) l++;
79             cmin(ans2, Abs(n - 2 * (b[r].id - b[l].id)));
80             if (l != i) 
81                 cmin(ans2, Abs(n - 2 * (b[r].id - b[l - 1].id)));
82         }
83     }
84     printf("%lld %d
", ans1, ans2);
85 }

Luogu3587 POD

Luogu3589[POI2015]KUR

神奇的传送门~

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原文地址：https://www.cnblogs.com/cychester/p/9820393.html