1427 - Substring Frequency (II)

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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.

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]]);
		}
	}
}