kmp 学习资料： [

](http://blog.csdn.net/v_july_v/article/details/7041827)

Number Sequence

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 32768/32768 K
(Java/Others)
Total Submission(s): 12419 Accepted Submission(s): 5661

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

Source

HDU 2007-Spring Programming Contest

kmp 模板题求字串在母串第一次出现的位置

#include<stdio.h>
#include<iostream>
#include<math.h>
#include<stdlib.h>
#include<ctype.h>
#include<algorithm>
#include<vector>
#include<string.h>
#include<queue>
#include<stack>
#include<set>
#include<map>
#include<sstream>
#include<time.h>
#include<malloc.h>

using namespace std;

int a[1000005],b[10005];
int next[10005];

void get_next(int b[],int m)
{
    int i=0;
    next[0]=-1;
    int j=-1;
    while(i<m)
    {
     if(j==-1||b[i]==b[j])
     {
        ++i;
        ++j;
      if(b[i]==b[j])
      next[i]=next[j];
      else
      next[i]=j;
    }
    else
    j=next[j];
   }
}
int KMP(int a[],int n,int b[],int m,int next[],int pos)
{
    int i=pos;
    int j=0;
    while(i<n&&j<m)
    {
    if(j==-1||a[i]==b[j])
    {
        ++i;
        ++j;
    }
    else
    j=next[j];
    }
   if(j>=m)
   return i-j+1;
   else
   return -1;
}
int main()
{
    int test,n,m,i,j;
    scanf("%d",&test);
    while(test--)
    {
     scanf("%d %d",&n,&m);
     for(i=0;i<n;++i)
     scanf("%d",&a[i]);
     for(j=0;j<m;++j)
     scanf("%d",&b[j]);
     get_next(b,m);
     cout<<KMP(a,n,b,m,next,0)<<endl;
    }
}