HDU 1711 Number Sequence KMP模板

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)

Total Submission(s): 26895    Accepted Submission(s): 11360

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

http://blog.csdn.net/v_july_v/article/details/7041827 这篇博客kmp写的很详细（大牛的博客）。

KMP的模板题：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>#define maxx 10100using namespace std;int a[maxx*100];int b[maxx];int nex[maxx];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;    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);        printf("%d\n",kmp());    }}