hdu 1711 Number Sequence

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Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 31372    Accepted Submission(s): 13170

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

KMP算法，得到跳转表（next数组），然后进行匹配。

#include <iostream>#include <cstring>#include <cstdio>const int M=1e4;const int N=1e6;int Next[M+10];int str[N+10];int mo[M+10];int n,m;using namespace std;void getNext(){Next[0]=-1;int i=0,j=-1;while(i<m){if(j==-1||mo[i]==mo[j]){i++;j++;Next[i]=j;}elsej=Next[j];}}int main(){int T;scanf("%d",&T);while(T--){scanf("%d%d",&n,&m);for(int i=0;i<n;i++)scanf("%d",&str[i]);for(int i=0;i<m;i++)scanf("%d",&mo[i]);getNext();int ans=-1;int i=0,j=0;while(i<n)//KMP匹配{if(j==-1||str[i]==mo[j]){i++;j++;}elsej=Next[j];if(j==m){ans=i-j+1;break;}}printf("%d\n",ans);}return 0;}