HDU 1711 KMP求匹配位置

Number Sequence

Time Limit:5000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

  HDU 1711

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 
     

KMP基础，求匹配位置：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
const int MAX=1000010;
int nex[MAX];
int a[MAX],b[MAX];
int len1,len2;
void get_next()
{
    int j=-1,i=0;
    nex[0]=-1;
    while(i<len2)
    {
        if(j==-1||b[i]==b[j])
        {
            i++;j++;
            nex[i]=j;
        }
        else
        j=nex[j];
    }
}
int kmp(){
    get_next();
    int i=0,j=0;
    while(i<len1){
        if(j==-1||a[i]==b[j]){
            i++;
            j++;
        }else
            j=nex[j];
        if(j==len2)
            return i-j+1;
    }
    return -1;
}
int main()
{
    int n;
    cin>>n;
    while(n--){
        cin>>len1>>len2;
        for(int i=0;i<len1;i++)
            scanf("%d",&a[i]);
        for(int i=0;i<len2;i++)
            scanf("%d",&b[i]);
        cout<<kmp()<<endl;
    }
}