题目链接：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

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1711

看了视频也看了博客，对KMP算法算是有了一知半解，先来一发水题吧。

对于next 数组 只需要求模式串，也就是两个串匹配中较小的那个。

【源代码】

#include <iostream>
#include <cstdio>
using namespace std;
const int maxn =1000005;
const int maxb =10005;
int a[maxn],b[maxb];
int nextv[maxb];
void get_next(int b[],int m){ //求模式串的 next数组
	int i=0;//前缀
	nextv[0]=-1;
	int j=-1;//后缀
	while(i<m){
		if(j==-1||b[i]==b[j]){
			i++; j++;
			if(b[i]==b[j]) //遇到相同元素的优化
				nextv[i]=nextv[j];
			else
				nextv[i]=j;
		}
		else
			j=nextv[j];
	}
}
int KMP(int n,int m){
	int i=0;
	int j=0;
	while(i<n&&j<m){
		if(j==-1||a[i]==b[j]){
			i++;j++;
		}
		else{
			j=nextv[j];
		}
	}
	if(j==m){ //只有匹配出结果j才能==m
		// cout<<"i: "<<i<<" "<<j<<endl;
		return i-m+1;
	}
	else return -1;
}
int main(){
	int T,n,m;
	scanf("%d",&T);
	while(T--){
		scanf("%d%d",&n,&m);
		for(int i=0;i<n;i++){
			scanf("%d",&a[i]);
		}
		for(int i=0;i<m;i++){
			scanf("%d",&b[i]);
		}
		get_next(b,m); //模式串 partern
		int ans=KMP(n,m);
		cout<<ans<<endl;
	}
	return 0;
}

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