题意：给定一个序列 s1, s2,...sn,以及一个k，求一个连续的k个数，把s[i]...s[i+k-1]变成一个数s'，使得sigma(|s[j]-s'|)(i<=j<=i+k-1)最小
思路：最近无聊切起poi。。顿感智商不够。。
这道题很容易想到把这些数变成中位数最优。。那么就可以用平衡树维护了。。
当然，直接用set维护也就够了。。不过为了练练代码。。我还是写了下splay。。
code:
1 /* 2 * Author: Yzcstc 3 * Created Time: 2014/11/8 19:03:57 4 * File Name: klo.cpp 5 */ 6 #include<cstdio> 7 #include<iostream> 8 #include<cstring> 9 #include<cstdlib> 10 #include<cmath> 11 #include<algorithm> 12 #include<string> 13 #include<map> 14 #include<set> 15 #include<vector> 16 #include<queue> 17 #include<stack> 18 #include<ctime> 19 #define repf(i, a, b) for (int i = (a); i <= (b); ++i) 20 #define M0(x) memset(x, 0, sizeof(x)) 21 #define Inf 0x7fffffff 22 using namespace std; 23 const int maxn = 120000; 24 typedef long long ll; 25 int n, k; 26 int a[101010]; 27 #define L ch[x][0] 28 #define R ch[x][1] 29 #define KT ch[ch[root][0]][1] 30 struct SplayTree{ 31 int sz[maxn], pre[maxn], ch[maxn][2], v[maxn]; 32 ll sum[maxn]; 33 int m, root; 34 void push_up(const int& x){ 35 sz[x] = sz[L] + sz[R] + 1; 36 sum[x] = sum[L] + sum[R] + v[x]; 37 } 38 void init(){ 39 M0(sz), M0(pre), M0(ch), M0(ch), M0(v); 40 } 41 void push_down(const int& x){ 42 } 43 void rotate(int &x, const int f){ 44 int y = pre[x], z = pre[y]; 45 push_down(y), push_down(x); 46 ch[y][!f] = ch[x][f]; 47 pre[ch[x][f]] = y; 48 pre[x] = pre[y]; 49 if (z) ch[z][ch[z][1] == y] = x; 50 ch[x][f] = y; 51 pre[y] = x; 52 push_up(y); 53 } 54 void splay(int &x, const int g){ 55 push_down(x); 56 while (pre[x] != g){ 57 int y = pre[x], z = pre[y]; 58 if (z == g) rotate(x, ch[y][0] == x); 59 else { 60 int f = (ch[z][0] == y); 61 ch[y][!f] ? rotate(y, f) : rotate(x, !f); 62 rotate(x, f); 63 } 64 } 65 push_up(x); 66 if (!g) root = x; 67 } 68 void rto(int k,const int g, int& y){ 69 int x = root; 70 while (sz[L] + 1 != k){ 71 push_down(x); 72 if (sz[L] >= k) x = L; 73 else { 74 k -= sz[L] + 1; 75 x = R; 76 } 77 } 78 y = x; 79 splay(x, g); 80 } 81 int search(int x, int val){ 82 int y = -1; 83 while (x){ 84 y = x; 85 if (val <= v[x]) y = x, x = L; 86 else y = x, x = R; 87 } 88 return y; 89 } 90 void insert(int x, int val){ 91 int y = search(root, val); 92 ch[y][val > v[y]] = x; 93 pre[x] = y; 94 while (y) push_up(y), y = pre[y]; 95 splay(x, 0); 96 } 97 int findpre(int x){ 98 x = L; 99 while (R) x = R; 100 return x; 101 } 102 void split(int x){ 103 splay(x, 0); 104 int lt = findpre(x); 105 if (lt){ 106 splay(lt, x); 107 root = lt, pre[root] = 0; 108 ch[root][1] = R; 109 if (R) pre[R] = root; 110 } else root = R, pre[R] = 0; 111 push_up(root); 112 L = R = 0; 113 } 114 ll query(const int& n, int &vv){ 115 int k = (n+1) / 2, x; 116 if (!(n & 1)) ++k; 117 rto(k, 0, x); 118 ll res = (ll)v[x] * (k-1) - sum[ch[root][0]] + sum[ch[root][1]] - (ll)v[x] * (n-k); 119 vv = v[x]; 120 return res; 121 } 122 } S; 123 124 void init(){ 125 for (int i = 1; i <= n; ++i) 126 scanf("%d", &a[i]); 127 } 128 129 void solve(){ 130 if (k == 1){ 131 puts("0"); 132 for (int i = 1; i <= n; ++i) 133 printf("%d ", a[i]); 134 return; 135 } 136 S.root = 1, S.sz[1] = 1, S.sum[1] = S.v[1] = a[1]; 137 for (int i = 2; i <= n; ++i) 138 S.sz[i] = 1, S.v[i] = S.sum[i] = a[i]; 139 for (int i = 2; i <= k; ++i) 140 S.insert(i, a[i]); 141 int ans_pos = 1, ans_val = 0, val; 142 ll ans = S.query(k, ans_val), tmp; 143 for (int i = k + 1; i <= n; ++i){ 144 S.split(i-k); 145 S.insert(i, a[i]); 146 tmp = S.query(k, val); 147 if (tmp < ans) 148 ans = tmp, ans_val = val, ans_pos = i - k + 1; 149 } 150 cout << ans << endl; 151 for (int i = ans_pos; i <= ans_pos+k-1; ++i) a[i] = ans_val; 152 for (int i = 1; i <= n; ++i) printf("%d ", a[i]); 153 } 154 155 int main(){ 156 // freopen("a.in", "r", stdin); 157 // freopen("a.out", "w", stdout); 158 while (scanf("%d%d", &n, &k) != EOF){ 159 init(); 160 solve(); 161 } 162 return 0; 163 }
题意：有 n(n<=1000)个串，每个串的程度都<=100，有 m 次操作，每次操作都是将第 p1个串的第 w1 个字母和第 p2 个串的第 w2 个字母交换，问对于每一个串 i，在这些操作执行的过程中，它最多几个串相同过。
思路：由于每个串只有100，我们可以用n*m的时间进行hash。。
那么剩下来就是直接维护hash值为某一个数的集合了。。。平衡树可维护。。
code：
1 /* 2 * Author: Yzcstc 3 * Created Time: 2014/11/8 13:53:39 4 * File Name: poc.cpp 5 */ 6 #include<cstdio> 7 #include<iostream> 8 #include<cstring> 9 #include<cstdlib> 10 #include<cmath> 11 #include<algorithm> 12 #include<string> 13 #include<map> 14 #include<set> 15 #include<vector> 16 #include<queue> 17 #include<stack> 18 #include<ctime> 19 #define M0(x) memset(x, 0, sizeof(x)) 20 #define Inf 0x7fffffff 21 using namespace std; 22 #define M 10003 23 const int maxn = 201000; 24 /*** splay-tree***/ 25 #define L ch[x][0] 26 #define R ch[x][1] 27 int sz[maxn], pre[maxn], rt[maxn], ms[maxn], ch[maxn][2], lz[maxn]; 28 29 void Splay_init(){ 30 M0(sz), M0(pre), M0(rt), M0(ms), M0(ch), M0(lz); 31 } 32 33 void pushdown(const int &x){ 34 if (lz[x]){ 35 if (L) lz[L] = max(lz[L], lz[x]), ms[L] = max(ms[L], lz[x]); 36 if (R) lz[R] = max(lz[R], lz[x]), ms[R] = max(ms[R], lz[x]); 37 lz[x] = 0; 38 } 39 } 40 41 void pushup(int x){} 42 43 void rotate(int &x, const int f){ 44 int y = pre[x], z = pre[y]; 45 pushdown(y), pushdown(x); 46 ch[y][!f] = ch[x][f]; 47 pre[ch[x][f]] = y; 48 pre[x] = pre[y]; 49 if (z) ch[z][ch[z][1] == y] = x; 50 ch[x][f] = y; 51 pre[y] = x; 52 pushup(y); 53 } 54 55 void splay(int &x, const int g, int &root){ 56 pushdown(x); 57 while (pre[x] != g){ 58 int y = pre[x], z = pre[y]; 59 if (z == g) rotate(x, ch[y][0] == x); 60 else { 61 int f = (ch[z][0] == y); 62 ch[y][!f] ? rotate(y, f) : rotate(x, !f); 63 rotate(x, f); 64 } 65 } 66 pushup(x); 67 if (!g) root = x; 68 } 69 70 void update(int &root, int s){ 71 lz[root] = max(lz[root], s); 72 ms[root] = max(ms[root], s); 73 } 74 75 int right(int x){ 76 while (R) x = R; 77 return x; 78 } 79 80 void insert(int& root, int x, int size){ 81 int p = right(root); 82 splay(p, 0, root); 83 ch[p][1] = x, pre[x] = p; 84 update(root, size); 85 } 86 87 void split(int& root, int x){ 88 splay(x, 0, root); 89 int lt = ch[x][0]; 90 lt = right(lt); 91 if (lt){ 92 splay(lt, x, root), root = lt; 93 if (R) pre[R] = root; 94 pre[root] = 0, ch[root][1] = R; 95 } else 96 root = R, pre[R] = 0; 97 L = R = 0; 98 } 99 100 /*** splay-end***/ 101 map<int, int> mp; 102 char s[1100][110]; 103 int hs[1010], n, m, q, pw[1010]; 104 105 inline int Hash(const char s[]){ 106 int res = 0; 107 for (int i = 0; i < m; ++i) 108 res = res * M + s[i]; 109 return res; 110 } 111 112 void init(){ 113 pw[m-1] = 1; 114 for (int i = m-2; i >= 0; --i) 115 pw[i] = pw[i+1] * M; 116 for (int i = 1; i <= n; ++i){ 117 scanf("%s", s[i]); 118 hs[i] = Hash(s[i]); 119 } 120 } 121 122 int f[maxn]; 123 void solve(){ 124 Splay_init(); 125 mp.clear(); 126 int tot = 0, u, v; 127 for (int i = 1; i <= n; ++i){ 128 u = hs[i], v = mp[u]; 129 if (v) 130 insert(rt[v], i, ++sz[v]); 131 else { 132 v = mp[u] = ++tot; 133 sz[v] = 1; rt[v] = i; 134 update(rt[v], 1); 135 } 136 } 137 int a, x1, b, x2; 138 int h1, h2; 139 while (q--){ 140 scanf("%d%d%d%d", &a, &x1, &b, &x2); 141 x1--, x2--; 142 if (a == b){ 143 h1 = hs[a] + pw[x1] * (s[b][x2] - s[a][x1]) + pw[x2] * (s[a][x1] - s[b][x2]); 144 u = mp[hs[a]], split(rt[u], a); 145 if (--sz[u] == 0) mp[hs[a]] = rt[u] = 0; 146 v = mp[h1]; 147 if (v) 148 insert(rt[v], a, ++sz[v]); 149 else { 150 v = mp[h1] = ++tot; 151 sz[v] = 1; rt[v] = a; 152 update(rt[v], 1); 153 } 154 hs[a] = h1; 155 swap(s[a][x1], s[b][x2]); 156 continue; 157 } 158 h1 = hs[a] + pw[x1] * (s[b][x2] - s[a][x1]); 159 h2 = hs[b] + pw[x2] * (s[a][x1] - s[b][x2]); 160 u = mp[hs[a]], split(rt[u], a); 161 if (--sz[u] == 0) mp[hs[a]] = rt[u] = 0; 162 u = mp[hs[b]], split(rt[u], b); 163 if (--sz[u] == 0) mp[hs[b]]= rt[u] = 0; 164 v = mp[h1]; 165 if (v) 166 insert(rt[v], a, ++sz[v]); 167 else { 168 v = mp[h1] = ++tot; 169 sz[v] = 1; rt[v] = a; 170 update(rt[v], 1); 171 } 172 v = mp[h2]; 173 if (v) 174 insert(rt[v], b, ++sz[v]); 175 else { 176 v = mp[h2] = ++tot; 177 sz[v] = 1; rt[v] = b; 178 update(rt[v], 1); 179 } 180 hs[a] = h1, hs[b] = h2; 181 swap(s[a][x1], s[b][x2]); 182 } 183 for (int i = 1; i <= n; ++i){ 184 u = mp[hs[i]]; 185 splay(i, 0, rt[u]); 186 f[i] = ms[i]; 187 } 188 for (int i = 1; i <= n; ++i) 189 printf("%d ", f[i]); 190 } 191 192 int main(){ 193 // freopen("a.in", "r", stdin); 194 // freopen("a.out", "w", stdout); 195 while (scanf("%d%d%d", &n, &m, &q) != EOF){ 196 init(); 197 solve(); 198 } 199 fclose(stdin); fclose(stdout); 200 return 0; 201 }