KMP入门（匹配）

Description

Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1], b[2], ...... , b[M] (1 <= M <= 10000, 1 <= N <= 1000000). Your task is to find a number K which make a[K] = b[1], a[K + 1] = b[2], ...... , a[K + M - 1] = b[M]. If there are more than one K exist, output the smallest one.

Input

The first line of input is a number T which indicate the number of cases. Each case contains three lines. The first line is two numbers N and M (1 <= M <= 10000, 1 <= N <= 1000000). The second line contains N integers which indicate a[1], a[2], ...... , a[N]. The third line contains M integers which indicate b[1], b[2], ...... , b[M]. All integers are in the range of [-1000000, 1000000].

Output

For each test case, you should output one line which only contain K described above. If no such K exists, output -1 instead.

Sample Input

2
13 5
1 2 1 2 3 1 2 3 1 3 2 1 2
1 2 3 1 3
13 5
1 2 1 2 3 1 2 3 1 3 2 1 2
1 2 3 2 1

Sample Output

6
-1

解题思路：题目大意：判断第一个串中是否存在第二个串，存在则输出最小的匹配起始位置，不存在则输出-1。
利用fail数组，KMP.

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

using namespace std;

int N[],M[],fail[];
int n,m;
void KMP();

int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&n,&m);
        for(int i=;i<=n;i++)
        scanf("%d",&N[i]);
        for(int i=;i<=m;i++)
        scanf("%d",&M[i]);
        KMP();
    }
    return ;
}

void KMP()
{
    memset(fail,,sizeof(fail));
    fail[]=-;//初始化fail数组
    for(int i=;i<=m;i++)
    {
        int p=fail[i-];
        //如果不满足要求则一直fail直到p=0
        while(p>=&&M[p+]!=M[i])
        p=fail[p];
        fail[i]=p+;
    }
    int ans=;
    for(int i=,p=;i<=n;i++)
    {
        while(p>=&&M[p+]!=N[i])
        p=fail[p];
        if(++p==m)
        {
            ans=i-m+;
            p=fail[p];
            break;
        }
    }
     if(ans) printf("%d\n",ans);
     else printf("-1\n");
}