HDOJ 1711 Number Sequence

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8600    Accepted Submission(s): 3953

Problem Description

Given two sequences of numbers : a[1], a[2], ...... , a[N], and b[1], b[2], ...... , b (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 . 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 . 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

HDU 2007-Spring Programming Contest

 

Recommend

lcy

 

 

没事水一把KMP。。。。。

#include <iostream> #include <cstdio> #include <cstring> using  namespace std; int n,m; int a[ 1000010],b[ 1000010],next[ 1000010]; int main() {      int t;     scanf( "%d",&t);      while(t--)     {         memset(next, 0, sizeof(next));         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);         }          for( int i= 1;i<m;i++)         {              int j=i;              while(j> 0)             {                 j=next[j];                  if(b ==b[j])                 {                     next[i+1]=j+1;                     break;                 }             }         }         int pos=-2;         for(int i=0,j=0;i<n;i++)         {             if(j<m&&a==b[j])             {                 j++;             }             else             {                 while(j>0)                 {                     j=next[j];                     if(a==b[j])                     {                         j++;                         break;                     }                 }             }             if(j==m)             {                 pos=i-m+1;                 break;             }         }         printf("%d\n",pos+1);     }     return 0; }

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