HDU 1711 Number Sequence（KMP）

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 25840    Accepted Submission(s): 10919

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

 

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KMP裸题，套板子即可

#include<bits/stdc++.h>
using namespace std;
const int N = 1000000 + 10;
int a[N];
int b[N];
int nex[N];
int la,lb;
void get_next()
{
    int i=0,j=-1;
    nex[0]=-1;
    while(i<lb)
    {
        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<la)
    {
        if(j==-1||a[i]==b[j])
        {
            i++;
            j++;
        }
        else
            j=nex[j];
        if(j==lb)
            return i-lb+1;
    }
    return -1;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d%d",&la,&lb);
        for(int i=0;i<la;i++)
            scanf("%d",&a[i]);
        for(int i=0;i<lb;i++)
            scanf("%d",&b[i]);
        printf("%d\n",kmp());
    }
}