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  • 【洛谷 P5357】 【模板】AC自动机(二次加强版)(AC自动机,差分)

    每次匹配都不停跳fail显然太慢了,于是在每个节点和fail指向的点连一条边,构成一棵树,在这棵树上差分一下就好了。
    AC自动机 就这个算法而言其实没用想象中那么难。

    #include <cstdio>
    #include <queue>
    #include <cstring>
    using namespace std;
    struct Node{
        int fail, next[26], num;
    }AC[200010];
    int n, u, cnt;
    queue <int> q;
    int p[200010], vis[200010];
    char a[2000010];
    int f[200010];
    struct Edge{
    	int next, to;
    }e[200010];
    int head[200010], num;
    inline void Add(int from, int to){
    	e[++num].to = to; e[num].next = head[from]; head[from] = num;
    }
    void insert(int x){
        int len = strlen(a + 1), w;
        u = 0;
        for(int i = 1; i <= len; ++i){
            w = a[i] - 'a';
            if(!AC[u].next[w])
              AC[u].next[w] = ++cnt;
            u = AC[u].next[w];
        }
        f[x] = u;
    }
    void BuildFail(){
    	u = 0;
        for(int i = 0; i < 26; ++i)
           if(AC[u].next[i])
             q.push(AC[u].next[i]);
        while(q.size()){
            u = q.front(); q.pop();
            for(int i = 0; i < 26; ++i)
               if(AC[u].next[i]){
               	 q.push(AC[u].next[i]);
                 AC[AC[u].next[i]].fail = AC[AC[u].fail].next[i];
               }
               else
    		     AC[u].next[i] = AC[AC[u].fail].next[i];
        }
    }
    void Match(){
        int len = strlen(a + 1);
        u = 0;
        for(int i = 1; i <= len; ++i){
            u = AC[u].next[a[i] - 'a'];
            ++vis[u];
        }
    }
    void dfs(int x){
    	for(int i = head[x]; i; i = e[i].next){
    		dfs(e[i].to); vis[x] += vis[e[i].to];
    	}
    }
    int main(){
        scanf("%d", &n);
        for(int i = 1; i <= n; ++i) f[i] = i;
        for(int i = 1; i <= n; ++i){
         	scanf("%s", a + 1);
        	insert(i);
        }
        scanf("%s", a + 1);
        BuildFail();
        Match();
        for(int i = 1; i <= cnt; ++i)
           Add(AC[i].fail, i);
        dfs(0);
        for(int i = 1; i <= n; ++i)
           printf("%d
    ", vis[f[i]]);
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/Qihoo360/p/10883618.html
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