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
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 "); 52 else 53 printf("%d ",ans+1); 54 } 55 return 0; 56 }