hdu1711 Number Sequence

 

Number Sequence

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

Total Submission(s) : 9   Accepted Submission(s) : 4

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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

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

找匹配的位置，C++交500MS，G++交1250MS。

#include <stdio.h>
int s[1000005],ss[10005];
int next[10005];

void getnext(int x) {
 int i=0,j=-1;
 next[0]=-1;
  while(i<x) {
   if(j==-1||ss[i]==ss[j]){
    i++;j++;
   // if(next[i]==next[j])
  //    next[i]=next[j];
  //  else
      next[i]=j;
   }
   else j=next[j];
  }
}

int kmp(int x,int y){
 int i=0,j=0;

 getnext(y);

 while(i<x){
  if(j==-1||s[i]==ss[j]){
   i++;j++;
  }
  else j=next[j];

  if(j==y) return  i-j+1;

 }
 return -1;
}


int main()
{
 int t;
 int i,n,m;
 scanf("%d",&t);
 while(t--) {
  scanf("%d%d",&n,&m);

  for(i=0;i<n;i++) scanf("%d",s+i);
  for(i=0;i<m;i++) scanf("%d",ss+i);

  int k=kmp(n,m);

  printf("%d/n",k);

 }
}