HDU 1711 Number Sequence (KMP)

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 22861    Accepted Submission(s): 9770

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

 

Source

HDU 2007-Spring Programming Contest

 

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第一次利用kmp写题。。。

刚开始求next函数值时用了next数组，然后就一直CP..

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;int n,m;int a[1000100],b[100100];  int ne[10010]; void getne(){int i,j;i=1;j=0;ne[1]=0;while(i<m){if(j==0||b[i]==b[j]){++i;++j;ne[i]=j;}elsej=ne[j];}}void KMP(){int i,j,ans;bool fafe=false; i=1;j=1;while(i<=n){        while (j>0&&a[i]!=b[j])              j=ne[j];          i++;j++;          if (j==m+1)          {              fafe=true;              ans=i-m;              break;          }  }    if (fafe)      printf("%d\n",ans);      else      printf("-1\n"); }int main(){int t,i,j;scanf("%d",&t);while(t--){scanf("%d%d",&n,&m);for(i=1;i<=n;i++){scanf("%d",&a[i]); }for(i=1;i<=m;i++){scanf("%d",&b[i]);}getne();KMP();}return 0;}