ZOJ 3587 Marlon's String

KMP，每匹配到一点就记录一下，然后用fail数组把已经匹配到的但是kmp没有记录的点加上。。。。

Marlon's String

Time Limit: 2000MS

Memory Limit: 65536KB

64bit IO Format: %lld & %llu

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Description

Long long ago, there was a coder named Marlon. One day he picked two string on the street. A problem suddenly crash his brain...

Let S_i..j denote the i-th character to the j-th character of string S.

Given two strings S and T. Return the amount of tetrad (a,b,c,d) which satisfy S_a..b + S_c..d = T , a≤b and c≤d.

The operator + means concate the two strings into one.

Input

The first line of the data is an integer Tc. Following Tc test cases, each contains two line. The first line is S. The second line is T. The length of S and T are both in range [1,100000]. There are only letters in string S and T.

Output

For each test cases, output a line for the result.

Sample Input

1
aaabbb
ab

Sample Output

9

Source

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#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn=200000;
int fail[maxn],n,m,c1[maxn],c2[maxn];
char S[maxn],T[maxn];

void getfail(char*T,int*f)
{
    f[0]=f[1]=0;
    for(int i=1;i<m;i++)
    {
        int j=f[i];
        while(j&&T[i]!=T[j]) j=fail[j];
        f[i+1]=(T[i]==T[j])? j+1 : 0;
    }
}

void kmp(char*S,char*T,int*f,int* c)
{
    getfail(T,f);
    int j=0;
    for(int i=0;i<n;i++)
    {
        while(j&&T[j]!=S[i]) j=f[j];
        if(T[j]==S[i])
        {
            j++;
            c[j]++;
        }
    }
    for(int i=m;i>=0;i--)
        if(f[i]) c[f[i]]+=c[i];
}

int main()
{
    int tk;
    scanf("%d",&tk);
while(tk--)
{
    scanf("%s%s",S,T);

    memset(c1,0,sizeof(c1));
    memset(c2,0,sizeof(c2));

    n=strlen(S);m=strlen(T);
    kmp(S,T,fail,c1);

    reverse(S,S+n); reverse(T,T+m);
    kmp(S,T,fail,c2);

    long long int ans=0;
    for(int i=0;i<m;i++)
    {
        ans+=(long long int)c1[i]*c2[m-i];
    }

    printf("%lld\n",ans);
}
    return 0;
}