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

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

AC代码：

#include<iostream> #include<string.h> #define N 10001 #define M 1000001 using namespace std; int a[M],b[N],next[N]; int n,m; void KMP_next() { int i=1,j=0; next[0]=0; while(i<m) { if(b[j]==b[i]) { next[i]=j+1; i++;j++; } else{ if(j>0) j=next[j-1]; else next[i++]=0; } } } int main() { int Case; cin>>Case; while(Case--) { bool flag=false; memset(next,0,sizeof(next)); cin>>n>>m; for(int i=0;i<n;i++) cin>>a[i]; for(int i=0;i<m;i++) cin>>b[i]; KMP_next(); int i=0,j=0; while(i<n) { if(a[i]==b[j]) { if(j==m-1) {flag=true;break;} j++;i++; } else { if(j>0) j=next[j-1]; else i++; } } if(flag) cout<<i-j+1<<endl; else cout<<"-1"<<endl; } return 0; }