原题http://acm.hdu.edu.cn/showproblem.php?pid=1711

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem 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

#include <stdio.h>
#include <string.h>
#include <algorithm>
int n,m,next[10010],a[1000010],b[10010];

void Getnext(){
	int i=0;
	int j=-1;
	next[0] = -1;
	while(i < m){
		if(j==-1 || b[i]==b[j]){
			i++;
			j++;
			next[i] = j;
		}
		else{
			j = next[j];
		}
	}
}

int KMP(){
	int i=0,j=0;
	while(i<n && j<m){
		if(j==-1 || a[i]==b[j]){
			i++;
			j++;
		}
		else{
			j = next[j];
		}
	}
	if(j == m){
		return i-j+1;
	}
	else{
		return 0;
	}
}

int main(){
	int t,i;

	while(~scanf("%d",&t)){
		while(t--){
			scanf("%d%d",&n,&m);
			for(i=0;i<n;i++){
				scanf("%d",&a[i]);
			}
			for(i=0;i<m;i++){
				scanf("%d",&b[i]);
			}
			Getnext();
			int ans = KMP();
			if(ans){
				printf("%d\n",ans);
			}
			else{
				printf("-1\n");
			}
		}
	}

	return 0;
}