Substring Frequency LightOJ - 1255

 转载请注明出处，谢谢。http://www.cnblogs.com/acmer-roney/---by Roney

Substring Frequency

Time Limit: 1000MS

Memory Limit: 65536KB

64bit IO Format: %lld & %llu

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Description

A string is a finite sequence of symbols that are chosen from an alphabet. In this problem you are given two non-empty strings A and B, both contain lower case English alphabets. You have to find the number of times B occurs as a substring of A.

Input

Input starts with an integer T (≤ 5), denoting the number of test cases.

Each case starts with two lines. First line contains A and second line contains B. You can assume than 1 ≤ length(A), length(B) ≤ 10⁶.

Output

For each case, print the case number and the number of times B occurs as a substring of A.

Sample Input

4

axbyczd

abc

abcabcabcabc

abc

aabacbaabbaaz

aab

aaaaaa

aa

Sample Output

Case 1: 0

Case 2: 4

Case 3: 2

Case 4: 5

题目连接： http://acm.hust.edu.cn:8080/judge/problem/viewProblem.action?id=26965

解题思路：做此题一定要对KMP的next[]数组有充分的理解。针对此题，当匹配成功后，对计数器ans加一，并当作匹配失败处理，即执行j=next[j]，继续和主串进行匹配。

注意：此题的next[]数组是用修正前的next[]数组！

 

AC代码：

#include<cstdio>
#include<cstring>
#define N 1000001
char s1[N],s2[N];
int next[N];
int len1,len2;
void getnext()
{
    int j=0,k=-1;
    next[0]=-1;
    len2=strlen(s2);
    while(j<len2){
        if(k==-1||s2[j]==s2[k]){
            j++;k++;
            if(s2[j]!=s2[k])next[j]=k;
            else next[j]=next[k];
        }
        else k=next[k];
    }
}
int kmp()
{
    getnext();
    int i=0,j=0,ans=0;
    len1=strlen(s1);
    while(i<len1){
        if(j==-1||s1[i]==s2[j]){
            i++;j++;
            if(j>=len2){
                j=next[j];
                ans++;
            }
        }
        else j=next[j];
    }
    return ans;
}
int main()
{
    //freopen("in.txt","r",stdin);
    int t,c=0;scanf("%d",&t);
    while(t--){
        scanf("%s %s",s1,s2);
        printf("Case %d: %d\n",++c,kmp());
    }
    return 0;
}