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  • HDU 2222 Keywords Search(AC自己主动机模板题)

    题意:给出一个字符串和若干个模板,求出在文本串中出现的模板个数。

    思路:由于有可能有反复的模板,trie树权值记录每一个模板出现的次数就可以。

    #include<cstdio>
    #include<cstring>
    #include<cmath>
    #include<cstdlib>
    #include<iostream>
    #include<algorithm>
    #include<vector>
    #include<map>
    #include<queue>
    #include<stack>
    #include<string>
    #include<map>
    #include<set>
    #include<ctime>
    #define eps 1e-6
    #define LL long long
    #define pii (pair<int, int>)
    //#pragma comment(linker, "/STACK:1024000000,1024000000")
    using namespace std;
    
    const int maxn = 1000000 + 100;
    const int SIGMA_SIZE = 26;
    const int maxnode = 1000000+100;
    int n, ans;
    bool vis[maxn];
    map<string, int> ms;
    int ch[maxnode][SIGMA_SIZE+5];  
    int val[maxnode]; 
    int idx(char c) {return c - 'a';}
    struct Trie {   
        int sz;  
        Trie() { sz = 1; memset(ch[0], 0, sizeof(ch[0])); memset(vis, 0, sizeof(vis)); }          
        void insert(char *s) {  
            int u = 0, n = strlen(s);  
            for(int i = 0; i < n; i++) {  
                int c = idx(s[i]);  
                if(!ch[u][c]) {  
                    memset(ch[sz], 0, sizeof(ch[sz]));  
                 	val[sz] = 0;
                    ch[u][c] = sz++;  
                }    
                u = ch[u][c];  
            }  
            val[u]++;
        }  
    };  
    
     
    //ac自己主动机
    int last[maxn], f[maxn];
    void print(int j) {
    	if(j && !vis[j]) {
    		ans += val[j]; vis[j] = 1;
    		print(last[j]);
    	}
    } 
    
    int getFail() {
    	queue<int> q;
    	f[0] = 0;
    	for(int c = 0; c < SIGMA_SIZE; c++) {
    		int u = ch[0][c];
    		if(u) {
    			f[u] = 0; q.push(u); last[u] = 0;
    		}
    	}
    	while(!q.empty()) {
    		int r = q.front(); q.pop();
    		for(int c = 0; c < SIGMA_SIZE; c++) {
    			int u = ch[r][c];
    			if(!u) {
    				ch[r][c] = ch[f[r]][c];
    				continue;
    			}
    			q.push(u);
    			int v = f[r];
    			while(v && !ch[v][c]) v = f[v];
    			f[u] = ch[v][c];
    			last[u] = val[f[u]] ?

    f[u] : last[f[u]]; } } } void find_T(char* T) { int n = strlen(T); int j = 0; for(int i = 0; i < n; i++) { int c = idx(T[i]); j = ch[j][c]; if(val[j]) print(j); else if(last[j]) print(last[j]); } } char tmp[105]; char text[1000000+1000]; int main() { //freopen("input.txt", "r", stdin); int T; cin >> T; while(T--) { scanf("%d", &n); Trie trie; ans = 0; for(int i = 0; i < n; i++) { scanf("%s", tmp); trie.insert(tmp); } getFail(); scanf("%s", text); find_T(text); cout << ans << endl; } return 0; }


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  • 原文地址:https://www.cnblogs.com/mthoutai/p/6964416.html
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