HDU 1711 Number Sequence

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11789    Accepted Submission(s): 5380

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

 

裸地KMP，字符匹配，求出第一个匹配数字的位置

#include <iostream>#include<stdio.h>#include<string>#include<cstring>#include<algorithm>#define N 1000010using namespace std;int f[N];int a[N],b[N];int n,m;int main(){    int ca;    while(~scanf("%d",&ca))    {        while(ca--)        {           scanf("%d %d",&n,&m);           for(int i=0;i<n;i++)           scanf("%d",&a[i]);           for(int i=0;i<m;i++)           scanf("%d",&b[i]);            int j=-1;            f[0]=-1;            for(int i=1;i<m;i++)            {                while(j>=0&&b[j+1]!=b[i])                   j=f[j];                   if(b[j+1]==b[i])                   j++;                   f[i]=j;            }            j=-1;            int ans=-1;            for(int i=0;i<n;i++)            {                while(j>=0&&b[j+1]!=a[i])                  j=f[j];                  if(b[j+1]==a[i])                  j++;                  if(j==m-1)                  {                      ans=i-m+1+1;                      break;                  }            }            if(ans==-1)            {                printf("-1\n");            }            else            printf("%d\n",ans);        }    }    return 0;}