HDU - 1711 Number Sequence

题目：

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.

InputThe 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].
OutputFor 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

很简单，KMP算法模板题
直接放代码：

#include<iostream>
using namespace std;

int N,M;

//计算next数组
int* get_next(int* b){
    int* next=new int[M];
    next[0]=-1;
    int k=-1;
    for(int q=1;q<M;q++){
        while(k>-1&&b[k+1]!=b[q]){
            k=next[k];
        }
        if(b[k+1]==b[q]){
            k++;
        }
        next[q]=k;
    }
    return next;
}

int main(){
    int T=0;
    cin >>T;
    for(int i=0;i<T;i++){
        cin >>N>>M;
        int *a=new int[N];
        int *b=new int[M];
        for(int j=0;j<N;j++){
            cin >>a[j];
        }
        for(int j=0;j<M;j++){
            cin >>b[j];
        }
        //KMP
        bool found=false;
        int* next=get_next(b);
        int k=-1;
        for(int j=0;j<N;j++){
            while(k>-1&&a[j]!=b[k+1]){
                k=next[k];
            }
            if(a[j]==b[k+1]){
                k++;
            }
            if(k==M-1){
                found=true;
                cout <<j-M+2<<endl;
                break;
            }
        }
        if(!found){
            cout <<-1<<endl;
        }
    }
    system("pause");
    return 0;
}