LightOJ 题目1427 - Substring Frequency (II)（AC自己主动机）

1427 - Substring Frequency (II)

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Time Limit: 5 second(s)

Memory Limit: 128 MB

A string is a finite sequence of symbols that are chosen from an alphabet. In this problem you are given a string T and n queries each with a string P_i, where the strings contain lower case English alphabets only. You have to find the number of times P_i occurs as a substring of T.

Input

Input starts with an integer T (≤ 10), denoting the number of test cases.

Each case starts with a line containing an integer n (1 ≤ n ≤ 500). The next line contains the string T (1 ≤ |T| ≤ 10⁶). Each of the next n lines contains a string P_i (1 ≤ |P_i| ≤ 500).

Output

For each case, print the case number in a single line first. Then for each string P_i, report the number of times it occurs as a substring of T in a single line.

Sample Input	Output for Sample Input
2 5 ababacbabc aba ba ac a abc 3 lightoj oj light lit	Case 1: 2 3 1 4 1 Case 2: 1 1 0

Notes

1.      Dataset is huge, use faster I/O methods.

2.       If S is a string then |S| denotes the length of S.

就是问子串在母串中的出现的次数（可重叠）

ac代码

#include<stdio.h>   
#include<string.h>   
#include<queue>   
#include<iostream>   
using namespace std;  
const int  maxnode=250*1000+10000;
const int sg_size=26;  
char str[1000100],key[550];
int pos[550];
struct Trie  
{  
    int ch[maxnode][sg_size];  
    int val[maxnode];     
    int f[maxnode];   
    int num[maxnode];
    int sz;  
    void init()
	{
		sz=1;
		memset(ch,0,sizeof(ch));
		memset(val,0,sizeof(val));
		memset(f,0,sizeof(f));
		memset(num,0,sizeof(num));
	}
    int idx(char c)  
    {  
        return c-'a';  
    }  
    int insert(char *s)  
    {  
        int u=0,i;  
        for(i=0;s[i];i++)  
        {  
            int c=idx(s[i]);  
            if(!ch[u][c])  
                ch[u][c]=sz++;;
            u=ch[u][c];  
        }  
        val[u]++;  
        num[u]=0;  
        return u;  
    } 
    void build_ac()  
    {  
        queue<int>q;  
        int i;  
        for(i=0;i<sg_size;i++)  
        {  
            if(ch[0][i])  
                q.push(ch[0][i]);  
        }  
        int r,c,u,v;  
        while(!q.empty())  
        {  
            r=q.front();  
            q.pop();  
            for(c=0;c<sg_size;c++)  
            {  
                u=ch[r][c];  
                if(!u)  
                    continue;  
                q.push(u);  
                v=f[r];  
                while(v&&ch[v][c]==0)  
                    v=f[v];      
                f[u]=ch[v][c];  
            }  
        }  
    }  
    void find(char *s)  
    { 
        int j=0;  
        for(int i=0;s[i];i++)  
        {  
            int c=idx(s[i]);  
            while(j&&ch[j][c]==0)  
                j=f[j];  
            j=ch[j][c];  
            int temp=j;  
            while(temp)  
            {  
                num[temp]++;  
                temp=f[temp];  
            }  
        }  
    } 
}ac;  
int main()
{
	int t,c=0;
	scanf("%d",&t);
	while(t--)
	{
		int n;
		scanf("%d",&n);	
		scanf("%s",str);
		int i;
		ac.init();
		for(i=1;i<=n;i++)
		{
			scanf("%s",key);
			pos[i]=ac.insert(key);
		}  
		ac.build_ac();
		ac.find(str);
		printf("Case %d:\n",++c);
		for(i=1;i<=n;i++)
		{
			printf("%d\n",ac.num[pos[i]]);
		}
	}
}