Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 25491    Accepted Submission(s): 10764


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
 题意：给你a b数组  在a串中定位b串 输出最小的匹配开始位置 没有则输出-1
 题解：kmp
 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int f[2000100];
 4 void get(int *p,int m)
 5 {
 6     int j=0;
 7     f[0]=f[1]=0;
 8     for(int i=1;i<m;i++)
 9     {
10         j=f[i];
11         while(j&&p[j]!=p[i]) j=f[j];
12         if(p[i]==p[j]) f[i+1]=j+1;
13         else f[i+1]=0;
14     }
15 }
16 int kmp(int *s,int *p,int n,int m)
17 {
18     int j=0;
19     int i;
20     for(i=0;i<n;i++)
21     {
22         while(j&&p[j]!=s[i]) j=f[j];
23         if(s[i]==p[j]) j++;
24         if(j==m)
25             break;
26     }
27     if(j==m)
28     return i-m+1;
29     else
30     return -1;
31 }
32 int p[20100],t[2000100];
33 int n,m;
34 int main()
35 {
36     int T;
37     scanf("%d",&T);
38     for(int k=1;k<=T;k++)
39     {
40         scanf("%d %d",&n,&m);
41         memset(t,0,sizeof(t));
42         memset(p,0,sizeof(p));
43         for(int i=0;i<n;i++)
44             scanf("%d",&t[i]);
45         for(int i=0;i<m;i++)
46             scanf("%d",&p[i]);
47         get(p,m);
48         int ans=0;
49         ans+=kmp(t,p,n,m);
50         if(ans==-1)
51             printf("-1\n");
52         else
53         printf("%d\n",ans+1);
54     }
55     return 0;
56 }