HDU 5918 kmp

Sequence I

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Problem Description

Mr. Frog has two sequences a1,a2,⋯,an and b1,b2,⋯,bm and a number p. He wants to know the number of positions q such that sequence b1,b2,⋯,bmis exactly the sequence aq,aq+p,aq+2p,⋯,aq+(m−1)p where q+(m−1)p≤n and q≥1.

 

Input

The first line contains only one integer T≤100, which indicates the number of test cases.

Each test case contains three lines.

The first line contains three space-separated integers 1≤n≤106,1≤m≤106 and 1≤p≤106.

The second line contains n integers a1,a2,⋯,an(1≤ai≤109).

the third line contains m integers b1,b2,⋯,bm(1≤bi≤109).

 

Output

For each test case, output one line “Case #x: y”, where x is the case number (starting from 1) and y is the number of valid q’s.

 

Sample Input

2
6 3 1
1 2 3 1 2 3
1 2 3
6 3 2
1 3 2 2 3 1
1 2 3

 

Sample Output

Case #1: 2
Case #2: 1

题意：给你序列a和序列b，整数p，问能找到几个q，使得aq,a(q+p)a(q+2*p)...a(q+(m-1)p)匹配上b。

题解：kmp预处理，重叠匹配即可。

#include<stdio.h>
#include<string.h>

int n,m,p,a[1000005],b[1000005];
int next[1000005],sum;

void getnext(){  
    int i,j;  
    next[0]=-1;
    i=0;j=-1;  
    while(i<m){
        if(j==-1||b[i]==b[j]){
            i++;j++;next[i]=j;  
        }   
        else j=next[j];  
    }
}   
  
void kmp(int k){  
    int i,j;  
    i=k;j=0;  
    while(i<n&&j<m){  
        if(j==-1||a[i]==b[j]){  
            i+=p;j++;  //每次跳跃p
        }
        else  
            j=next[j];  
    }  
    if(j==m){sum++;kmp(i-(m-1)*p);}//匹配成功  重叠匹配
}

int main()
{
    int t,cas=0;
    scanf("%d",&t);
    while(t--){
        sum=0;
        scanf("%d%d%d",&n,&m,&p);
        for(int i=0;i<n;i++)scanf("%d",&a[i]);
        for(int i=0;i<m;i++)scanf("%d",&b[i]);
        getnext();
        for(int i=0;i<p;i++)kmp(i);
        printf("Case #%d: %d\n",++cas,sum);
    }
    return 0;
}