正题

ybtoj AC自动机-3

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题目大意

给你一个字符串和若干匹配串，问你匹配串的前缀和字符串的最大匹配

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解题思路

先把所有匹配串丢进AC自动机，然后拿字符串去跑

每次只在当前位置存下贡献，然后按bfs的倒叙传递贡献，最后再倒着跑每个匹配串

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代码

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define ll long long
#define N 10000010
using namespace std;
int n, w, hd, tl, len, q[N], b[N], d[N], v[N], nx[N], fa[N], to[N][4];
char s[N], ss[N];
int check(char g)
{if (g == 'E') return 0;if (g == 'S') return 1;if (g == 'W') return 2;if (g == 'N') return 3;
}
int insert(char* s)
{int now = 0, n = strlen(s+1);for (int i = 1; i <= n; ++i){int y = check(s[i]);if (!to[now][y]) to[now][y] = ++w;fa[to[now][y]] = now;q[to[now][y]] = q[now] + 1;now = to[now][y];}return now;
}
void bfs()
{for (int i = 0; i < 4; ++i)if (to[0][i]) d[++tl] = to[0][i];while(hd < tl){int h = d[++hd];for (int i = 0; i < 4; ++i)if (!to[h][i]) to[h][i] = to[nx[h]][i];else nx[to[h][i]] = to[nx[h]][i], d[++tl] = to[h][i];}return;
}
void ask(char* s)
{int now = 0;for (int i = 1; i <= len; ++i){int y = check(s[i]);now = to[now][y];b[now] = 1;}return;
}
int find(int x)//倒着跑匹配串
{while(x && !b[x]) x = fa[x];return q[x];
}
int main()
{scanf("%d%d", &len, &n);scanf("%s", ss+1);for (int i = 1; i <= n; ++i){scanf("%s", s+1);v[i] = insert(s);}bfs();ask(ss);for (int i = tl; i > 0; --i)b[nx[d[i]]] |= b[d[i]];//传递贡献for (int i = 1; i <= n; ++i)printf("%d\n", find(v[i]));return 0;
}