ZOJ 题目3587 Marlon's String（KMP）

Marlon's String

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Time Limit: 2 Seconds     
Memory Limit: 65536 KB

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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

题目大意：Return the amount of tetrad (a,b,c,d) which satisfy
S_a..b + S_c..d = T , a≤b and
c≤d.

ac代码

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
char a[200010],b[200010];
int c1[200010],c2[200010];
int next[200010];
void getnext(char *a)
{
	next[0]=next[1]=0;
	int i;
	int m=strlen(a);
	for(i=1;i<m;i++)
	{
		int j=next[i];
		while(j&&a[i]!=a[j])
			j=next[j];
		next[i+1]=(a[i]==a[j])?j+1:0;
	}
}
void kmp(char *a,char *b,int *c)
{
	getnext(b);
	int j=0,i;
	int n=strlen(a);
	int m=strlen(b);
	for(i=0;i<n;i++)
	{
		while(j&&a[i]!=b[j])
			j=next[j];
		if(a[i]==b[j])
		{
			j++;
			c[j]++;
		}
	}
	for(i=m;i>=0;i--)
		if(next[i])
			c[next[i]]+=c[i];
}
int main()
{
	int t;
	scanf("%d",&t);
	while(t--)
	{
		int i;
		memset(c1,0,sizeof(c1));
		memset(c2,0,sizeof(c2));
		scanf("%s%s",a,b);
		int len1=strlen(a);
		int len2=strlen(b);
		kmp(a,b,c1);
		reverse(a,a+len1);
		reverse(b,b+len2);
		kmp(a,b,c2);
		long long ans=0;
		for(i=0;i<len2;i++)
			ans+=(long long)c1[i]*c2[len2-i];
		printf("%lld\n",ans);
	}
}