HDU 5384 Danganronpa (AC自己主动机模板题)

题意:给出n个文本和m个模板,求每一个文本中全部模板出现的总次数。

思路:Trie树权值记录每一个模板的个数。对于每一个文本跑一边find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<ctime>
#include<set>
#define eps 1e-6
#define LL long long
#define pii (pair<int, int>)
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;

//const int INF = 0x3f3f3f3f;
int n, k, shit;


const int maxn = 1000000 + 100;
const int SIGMA_SIZE = 26;
const int maxnode = 1000000+100;
int ch[maxnode][SIGMA_SIZE+5];  
int val[maxnode], cnt[100000+1000]; 
int idx(char c) {return c - 'a';}
struct Trie {   
    int sz;  
    Trie() { sz = 1; memset(ch[0], 0, sizeof(ch[0]));};          
    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 i, int j) {
	if(j) {
		shit += val[j]; //cout << i << endl;
		print(i, 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 len = strlen(T); //cout << len << endl; int j = 0; for(int i = 0; i < len; i++) { int c = idx(T[i]); j = ch[j][c]; if(val[j]) print(i, j); else if(last[j]) print(i, last[j]); } } string text[100000+100]; char tmp[10050], tt[105000]; int ans[100000+10000]; int main() { //freopen("input.txt", "r", stdin); int T; cin >> T; while(T--) { scanf("%d%d", &n, &k); memset(ans, 0, sizeof(ans[0])*(n+10)); Trie trie; for(int i = 0; i < n; i++) { scanf("%s", tmp); text[i] = tmp; } for(int i = 0; i < k; i++) { scanf("%s", tmp); //cout << tmp << endl; trie.insert(tmp); } getFail(); for(int i = 0; i < n; i++) { shit = 0; find_T((char*)text[i].c_str()); ans[i] = shit; } for(int i = 0; i < n; i++) printf("%d ", ans[i]); } return 0; }


原文地址:https://www.cnblogs.com/yutingliuyl/p/6952894.html