HDU 1711 Number Sequence(字符串匹配)

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10571    Accepted Submission(s): 4814

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

 

 

 

#include<iostream>
#include<cstdio>
using namespace std;
const int maxn1=1000010;
const int maxn2=10010;
int n,m;
int a[maxn1];
int b[maxn2];
int next[maxn2];
int main()
{
 int kmp (int *a,int *b,int *next);
 int t,i;
 cin>>t;
 while(t--)
 {
  cin>>n>>m;
  for(i=0;i<n;i++)
  {scanf("%d",&a[i]);}
  for(i=0;i<m;i++)
  {scanf("%d",&b[i]);}
  int ans=kmp(a,b,next);
  cout<<ans<<endl;
 }
 return 0;
}

//此函数用来求匹配串s串的next数组
void getnext (int *s,int *next)
{
    next[0]=next[1]=0;
    for (int i=1;i<m;i++)//m为匹配串s的长度
 {
        int j=next[i];
        while (j&&s[i]!=s[j])
            j=next[j];
        next[i+1]=s[i]==s[j]?

j+1:0;
    }
}
//此函数用来求匹配位置的值的函数。匹配不成功返回值-1
int kmp (int *a,int *b,int *next)
{
    getnext (b,next);/////////
    int j=0;
    for (int i=0;i<n;i++)
 {/////n为串1的长度
        while (j&&a[i]!=b[j])
            j=next[j];
        if (a[i]==b[j])
            j++;
        if (j==m)//m为串2的长度
            return i-m+2;
    }
    return -1;
}