历程
被暴打了
原因是钻进了 \(T4\) 的坑中。。。
先看完题，发现 \(T4\) 比较有意思，\(T2\) 没有想法
\(T3\) 挺容易，做法似乎很好想
\(T1\) 送分，十几分钟搞定
然后开 \(T4\)，推了一下，明白题目需要我们干的事情
开码，发现修改与标记有些烦，然后标记打错了，一直调到 \(11\) 点
疯了！
还有 \(\text{45 min}\)，只要忍痛弃了
看看 \(T3\)，很好做，\(5 min\) 打完
有充满自信，放弃 \(T2\) 部分分，继续调 \(T4\)
但始终没有意识到一处标记的问题
导致比赛结束被暴打（更重要的是，暴力每题普遍 \(80，90pts\) ！！神仙数据。。。）
\(\text{T1 1782. Travel}\)
这不 \(Floyd\) 预处理后分层最短路 \(spfa\) 啊！
正好复习最短路模板
$\text{Code}$
\(\text{Code}\)
#include <cstdio>
#include <queue>
#include <cstring>
#include <iostream>
#define RE register
using namespace std;
const int N = 105, INF = 0x3f3f3f3f;
int n, m, q, G[N][N], vis[N][N], bz[N][15], f[N][15];
struct node{int x, k;};
queue<node> Q;
void SPFA()
{
memset(f, INF, sizeof f), Q.push(node{1, 0}), f[1][0] = 0, bz[1][0] = 1;
while (!Q.empty())
{
node z = Q.front(); Q.pop();
for(RE int i = 1; i <= n; i++)
if (vis[z.x][i] && i ^ z.x)
{
int j = i, k = z.k, w = G[z.x][i];
if (!vis[i][z.x]) w *= 2, ++k;
if (k <= q && f[j][k] > f[z.x][z.k] + w)
{
f[j][k] = f[z.x][z.k] + w;
if (!bz[j][k]) Q.push(node{j, k}), bz[j][k] = 1;
}
}
bz[z.x][z.k] = 0;
}
}
int main()
{
scanf("%d%d%d", &n, &m, &q), memset(G, INF, sizeof G);
for(RE int i = 1, x, y, z; i <= m; i++)
scanf("%d%d%d", &x, &y, &z), G[x][y] = min(G[x][y], z), vis[x][y] = 1;
for(RE int k = 1; k <= n; k++)
for(RE int i = 1; i <= n; i++)
if (i ^ k)
for(RE int j = 1; j <= n; j++)
if (j ^ i && j ^ k)
vis[i][j] = (vis[i][j] | (vis[i][k] && vis[k][j]));
SPFA();
int ans = INF;
for(RE int i = 1; i <= q; i++) ans = min(ans, f[n][i]);
if (ans == INF) ans = -1;
printf("%d\n", ans);
}
\(\text{T2 3337. wyl8899的TLE}\)
思考答案的产生
从 \(A\) 的首位， \(B\) 的某个位置开始，求 \(LCP\)
然后魔法地同时跳过一格（计入贡献），再求 \(LCP\)
答案取最大值即可
也就是说可以枚举 \(B\) 的开始位置，二分加哈希求 \(LCP\) 即可
思维真不高。。。
也算复习字符串基操吧
$\text{Code}$
\(\text{Code}\)
#include <cstdio>
#include <cstring>
#include <iostream>
#define RE register
#define IN inline
using namespace std;
typedef long long LL;
const int N = 5e4 + 5, P1 = 1e9 + 7, P2 = 1e9 + 9, B = 29;
char s1[N], s2[N];
LL b1[N], b2[N];
struct Hash{
int n; LL p1[N], p2[N];
IN void init(char *s)
{
n = strlen(s + 1);
for(RE int i = 1; i <= n; i++)
p1[i] = (p1[i - 1] * B + s[i] - 'a') % P1, p2[i] = (p2[i - 1] * B + s[i] - 'a') % P2;
}
IN int get1(int l, int r){return (p1[r] - p1[l - 1] * b1[r - l + 1] % P1 + P1) % P1;}
IN int get2(int l, int r){return (p2[r] - p2[l - 1] * b2[r - l + 1] % P2 + P2) % P2;}
}h1, h2;
IN int lcp(int l1, int l2)
{
int l = 0, r = min(h1.n - l1 + 1, h2.n - l2 + 1), mid = l + r >> 1, res = 0;
for(; l <= r; mid = l + r >> 1)
if ((h1.get1(l1, l1 + mid - 1) == h2.get1(l2, l2 + mid - 1))
&& (h1.get2(l1, l1 + mid - 1) == h2.get2(l2, l2 + mid - 1))) res = mid, l = mid + 1;
else r = mid - 1;
return res;
}
int main()
{
b1[0] = b2[0] = 1;
for(RE int i = 1; i <= N - 3; i++) b1[i] = b1[i - 1] * B % P1, b2[i] = b2[i - 1] * B % P2;
scanf("%s%s", s1 + 1, s2 + 1), h1.init(s1), h2.init(s2);
int ans = 0;
for(RE int i = 1, len; i <= h2.n; i++)
ans = max(ans, (len = lcp(1, i)) + lcp(len + 2, i + len + 1) + (len < h1.n && i + len - 1 < h2.n));
printf("%d\n", ans);
}
\(\text{T3 3338. 法法塔的奖励}\)
考虑一个点的答案，显然由子树符合条件的 \(dp\) 值转来
发现是维护子树的问题，所以树上启发式合并加树状数组做到 \(O(n\log^2 n)\) 或者线段树合并做到 \(O(n\log n)\)
当然是打前者啦，跟打暴力一样一样的
$\text{Code}$
\(\text{Code}\)
#include <cstdio>
#include <iostream>
#define RE register
#define IN inline
using namespace std;
const int N = 1e5 + 5;
int n, fa[N], a[N], h[N], tot, c[N];
int dfc, siz[N], dfn[N], rev[N], son[N], ans[N];
struct edge{int to, nxt;}e[N];
IN void add(int x, int y){e[++tot] = edge{y, h[x]}, h[x] = tot;}
void dfs1(int x)
{
siz[x] = 1, dfn[x] = ++dfc, rev[dfc] = x;
for(RE int i = h[x]; i; i = e[i].nxt)
{
int v = e[i].to;
dfs1(v), siz[x] += siz[v];
if (siz[v] > siz[son[x]]) son[x] = v;
}
}
IN int lowbit(int x){return x & (-x);}
IN void update(int x, int v){for(; x <= n; x += lowbit(x)) c[x] = max(c[x], v);}
IN void clear(int x){for(; x <= n; x += lowbit(x)) c[x] = 0;}
IN int query(int x)
{
int res = 0;
for(; x; x -= lowbit(x)) res = max(res, c[x]);
return res;
}
void dfs2(int x, int kp)
{
for(RE int i = h[x]; i; i = e[i].nxt)
{
int v = e[i].to;
if (v == son[x]) continue;
dfs2(v, 0);
}
if (son[x]) dfs2(son[x], 1);
for(RE int i = h[x]; i; i = e[i].nxt)
{
int v = e[i].to;
if (v == son[x]){update(a[son[x]], ans[son[x]]); continue;}
for(RE int j = dfn[v]; j <= dfn[v] + siz[v] - 1; j++) update(a[rev[j]], ans[rev[j]]);
}
ans[x] = query(a[x]) + 1;
if (!kp) for(RE int i = dfn[x] + 1; i <= dfn[x] + siz[x] - 1; i++) clear(a[rev[i]]);
}
int main()
{
scanf("%d", &n); int x; scanf("%d", &x);
for(RE int i = 2; i <= n; i++) scanf("%d", &fa[i]), add(fa[i], i);
for(RE int i = 1; i <= n; i++) scanf("%d", &a[i]);
dfs1(1), dfs2(1, 1);
for(RE int i = 1; i <= n; i++) printf("%d ", ans[i]);
}
\(\text{T4 3339. wyl8899和法法塔的游戏}\)
很容易想到 \(nim\) 游戏的结论：异或和为零为先手必败态
那么先求异或和，需要第一步取完后留给对手必败态，即 \((a_r-x) \oplus \text{others} = 0\)
需 \(x=a_r-\text{others}\) 最大化，可持久化 \(Trie\) 即可
但是需要支持修改操作，所以考虑分块加 \(Trie\)
注意打标记时的细节
$\text{Code}$
\(\text{Code}\)
#include <cstdio>
#include <cmath>
#include <iostream>
#define IN inline
#define RE register
using namespace std;
const int N = 1e5 + 5, B = 319;
int n, m;
struct DS{
int pos[N], block, a[N], tag[B], tr[N * 20][2], size, sum[N * 20], rt[B];
IN void init()
{
block = sqrt(n);
for(RE int i = 1; i <= n; i++) pos[i] = (i - 1) / block + 1;
}
IN void update(int p, int x, int v)
{
if (!rt[p]) rt[p] = ++size;
int u = rt[p];
for(RE int i = 10; i >= 0; i--)
{
int ch = (x >> i) & 1;
if (!tr[u][ch]) tr[u][ch] = ++size;
sum[u = tr[u][ch]] += v;
}
}
IN int query(int p, int v)
{
int u = rt[p], res = 0;
for(RE int i = 10; i >= 0; i--)
{
int ch = (v >> i) & 1;
if (sum[tr[u][ch]]) u = tr[u][ch];
else if (sum[tr[u][ch ^ 1]]) u = tr[u][ch ^ 1], res |= (1 << i);
}
return res;
}
IN int get(int x){return a[x] ^ tag[pos[x]] ^ a[x + 1] ^ tag[pos[x + 1]];}
IN void Modify(int r, int v)
{
int t = get(r), w = a[r] ^ (t - v) ^ tag[pos[r]] ^ a[r + 1] ^ tag[pos[r + 1]];
for(RE int i = r; i >= (pos[r] - 1) * block + 1; i--)
update(pos[r], a[i], -1), update(pos[r], a[i] ^= w, 1);
for(RE int i = 1; i < pos[r]; i++) tag[i] ^= w;
}
IN int Query(int l, int r, int v)
{
int res = N;
for(RE int i = l; i <= min(pos[l] * block, r); i++) res = min(res, a[i] ^ tag[pos[l]] ^ v);
if (pos[l] ^ pos[r])
for(RE int i = (pos[r] - 1) * block + 1; i <= r; i++) res = min(res, a[i] ^ tag[pos[r]] ^ v);
for(RE int i = pos[l] + 1; i < pos[r]; i++) res = min(res, query(i, v ^ tag[i]));
return res;
}
}T;
int main()
{
scanf("%d", &n);
for(RE int i = 1; i <= n; i++) scanf("%d", &T.a[i]);
T.init();
for(RE int i = n; i; i--) T.a[i] ^= T.a[i + 1], T.update(T.pos[i], T.a[i], 1);
scanf("%d", &m);
for(RE int i = 1, ed, L, R; i <= m; i++)
{
scanf("%d%d%d", &ed, &L, &R);
int ans = T.get(ed) - T.Query(L, R, T.a[ed] ^ T.tag[T.pos[ed]]);
if (ans <= 0) ans = -1;
if (ans != -1) T.Modify(ed, ans);
printf("%d\n", ans);
}
}
可见这套题挺简单的，但是考场要分配好时间，避免陷入一处坑很久导致意外或者“正常”死亡