Number Sequence +KMP

Number Sequence

Time Limit : 10000/5000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 20   Accepted Submission(s) : 14

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
 
 
/*题目大意：给你两组数据，判断第二组是否存在于第一组，如果存在从第几位数开始，不存在则输出-1。思路：直接用KMP算法，找到第一个匹配的地方，然后将i-m+1及时第一次匹配的位置*/#include<iostream>#include<cstdio>using namespace std ;#define MAX 1000005 int next[MAX],m,n;int a[MAX] , b[MAX];void pre_next(int k ){int i = 0 , j = -1 ;next[0] = - 1 ;while( i < k ){if( j == -1 || a[i] == a[j]){i++ ;j++ ;next[i] = j ;}elsej = next[j] ;}}int  KMP(){int i = 0 , j = 0 ,count = 0 ;while( i <= n ){if(j== - 1 || a[j]==b[i]){i++;j++ ;}elsej = next[j] ;if(j == m ){return  i-m+1;}}return -1 ;}int main(void){int t ;scanf("%d",&t);while(t--){int i ;scanf("%d %d",&n,&m);for( i = 0 ; i < n ; i ++)scanf("%d",&b[i]);for( i = 0 ; i < m ; i ++)scanf("%d",&a[i]);pre_next(m);if(n<m)printf("-1\n");elseprintf("%d\n",KMP());}return 0 ; }