HDU1711

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 14563    Accepted Submission(s): 6392

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

 

Source

HDU 2007-Spring Programming Contest

 

Recommend

lcy

 

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;
int s[],t[];
int Next[];
int ans;
void getNext(int b){
   int i=,j=-;
   Next[]=-;
   while(i<b){
      if(j==-||t[i]==t[j]){
        i++;
        j++;
        Next[i]=j;
      }
    else
        j=Next[j];
   }
}
int kmp(int a,int b){
    int i=,j=,sum=;
    while(i<a){
        if(j==-||s[i]==t[j]){
            i++;
            j++;
        }
        else
            j=Next[j];
        if(j==b){
            ans=i+-b;///该步需要特别注意一下即可
            return ans;
        }
    }
    return -;
}
int main(){
   int tt;
   scanf("%d",&tt);
   while(tt--){
      memset(s,,sizeof(s));
      memset(t,,sizeof(t));
      memset(Next,,sizeof(Next));
      int t1,t2;
      scanf("%d%d",&t1,&t2);
     int temp;
      for(int i=;i<t1;i++){
         scanf("%d",&s[i]);
      }
           for(int i=;i<t2;i++){
           scanf("%d",&t[i]);
           }
      getNext(t2);
      int pd;
      pd=kmp(t1,t2);
      printf("%d\n",pd);

   }
   return ;
}