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因为BZOJ上“ 数据文件太过巨大,仅提供前三组数据测试.”所以我考场上写的60分的点分治交上去也A了。
我的这个点分治的时间复杂度是$O(Tnmlogn)$的,听题解时没听懂$O(Tnlogn)$的标算,还有听说标算要用到字符串哈希,然而我并不会,所以先留个坑,贴上自己的60分代码,满分做法等我学会哈希之后再做
#include<cstdio> #include<cstring> #include<algorithm> using namespace std; const int N = 1000003; void read(int &k) { k = 0; int fh = 1; char c = getchar(); for(; c < ‘0‘ || c > ‘9‘; c = getchar()) if (c == ‘-‘) fh = -1; for(; c >= ‘0‘ && c <= ‘9‘; c = getchar()) k = (k << 1) + (k << 3) + c - ‘0‘; k = k * fh; } struct node {int nxt, to;} E[N << 1]; int n, m, cnt, point[N], root, qu[N], fa[N], t[N], sz[N]; char c[N], mu[N]; bool vis[N], bo[N]; void ins(int x, int y) {E[++cnt].nxt = point[x]; E[cnt].to = y; point[x] = cnt;} void findrt(int x) { int p = 0, u, q = 1; qu[1] = x; fa[x] = 0; while (p != q) { u = qu[++p]; bo[u] = 1; sz[u] = 1; for(int tmp = point[u]; tmp; tmp = E[tmp].nxt) if (!vis[E[tmp].to] && E[tmp].to != fa[u]) qu[++q] = E[tmp].to, fa[E[tmp].to] = u; } for(int i = q; i >= 1; --i) { u = qu[i]; if (bo[u] && sz[u] * 2 > q) {root = u; return;} sz[fa[u]] += sz[u]; if (sz[u] * 2 > q) bo[fa[u]] = 0; } } int leftnow, rightnow, leftsum, rightsum, ret; void BFSleft(int x) { int p = 0, u, q = 1, tt; qu[1] = x; while (p != q) { u = qu[++p]; for(int tmp = point[u]; tmp; tmp = E[tmp].nxt) if (!vis[E[tmp].to] && E[tmp].to != fa[u] && c[E[tmp].to] == mu[tt = ((t[u] - 1 + m) % m)]) fa[E[tmp].to] = u, t[E[tmp].to] = tt, qu[++q] = E[tmp].to; } for(int i = 1; i <= q; ++i) if (t[qu[i]] == 0) ++leftnow; } void BFSright(int x) { int p = 0, u, q = 1, tt; qu[1] = x; while (p != q) { u = qu[++p]; for(int tmp = point[u]; tmp; tmp = E[tmp].nxt) if (!vis[E[tmp].to] && E[tmp].to != fa[u] && c[E[tmp].to] == mu[tt = ((t[u] + 1) % m)]) fa[E[tmp].to] = u, t[E[tmp].to] = tt, qu[++q] = E[tmp].to; } for(int i = 1; i <= q; ++i) if (t[qu[i]] == (m - 1)) ++rightnow; } void work(int x) { vis[x] = 1; int tt; for(int to = 0; to < m; ++to) if (c[x] == mu[to]) { leftsum = 0; rightsum = 0; if (to == 0) leftsum = 1; if (to == m - 1) rightsum = 1; for(int i = point[x]; i; i = E[i].nxt) if (!vis[E[i].to]) { leftnow = 0; if (c[E[i].to] == mu[tt = (to - 1 + m) % m]) fa[E[i].to] = x, t[E[i].to] = tt, BFSleft(E[i].to); rightnow = 0; if (c[E[i].to] == mu[tt = (to + 1) % m]) fa[E[i].to] = x, t[E[i].to] = tt, BFSright(E[i].to); ret += leftsum * rightnow; ret += rightsum * leftnow; leftsum += leftnow; rightsum += rightnow; leftnow = 0; rightnow = 0; } } for(int i = point[x]; i; i = E[i].nxt) if (!vis[E[i].to]) {findrt(E[i].to); work(root);} } int main() { freopen("pattern.in", "r", stdin); freopen("pattern.out", "w", stdout); int T, u, v; read(T); while (T--) { read(n); read(m); scanf("%s", c + 1); cnt = 0; memset(point, 0, sizeof(point)); memset(vis, 0, sizeof(vis)); for(int i = 1; i < n; ++i) {read(u); read(v); ins(u, v); ins(v, u);} scanf("%s", mu); ret = 0; findrt(1); work(root); printf("%d\n", ret); } return 0; }
【BZOJ 4598】【SDOI 2016 Round2 Day1 T3】模式字符串
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原文地址:http://www.cnblogs.com/abclzr/p/5514157.html