（HDU）1711

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 24863    Accepted Submission(s): 10530

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

213 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 1 313 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 2 1

 

Sample Output

6-1

 

KMP算法的入门题目，题意为输入一个原始串和一个模板串，在原始串中找到一个连续的和模板串一样的子串，再输出其开头索引，若没有就输出-1。注意KMP停止的时候索引的位置是匹配后模版串尾端的索引，所以要减去模版串长度后+1。

#include <bits/stdc++.h>using namespace std;const int MAXN=1000010;const int MAXM=10010;int a[MAXN],b[MAXM],nex[MAXM];int n,m;void preKMP(){    int p=0,q=-1; nex[0]=-1;    while(p<m){        while(q!=-1 && b[p]!=b[q]) q=nex[q];        nex[++p]=++q;    }}int KMP(){    preKMP();    int p=0,q=0;    while(p<n&&q<m){        while(q!=-1 && a[p]!=b[q]) q=nex[q];        p++; q++;        if(q==m) return p-m+1;    }    return -1;}int main(){    int t;    for(scanf("%d",&t);t;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]);        printf("%d\n",KMP());    }    return 0;}