hdu1711

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3595    Accepted Submission(s): 1637

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

213 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 1 313 51 2 1 2 3 1 2 3 1 3 2 1 21 2 3 2 1

 

Sample Output

6-1

 

Source

HDU 2007-Spring Programming Contest

 

Recommend

lcy

 

这个题目是kmp问题，关键在于构造失败函数。

#include<stdio.h>

#include<iostream>

using namespace std;

int s[1000000+2];

int t[10000+2];

int f[10000+2];

int n,m;

void fail()

{

    f[0]=-1;

    for(int j=1;j<m;j++)

    {

        int i=f[j-1];

        while(t[j]!=t[i+1]&&i>=0)i=f[i];

        if(t[j]==t[i+1])f[j]=i+1;

        else f[j]=-1;

    }

}

int kmp()

{

    fail();

    int i,j;

    i=0;

    j=0;

    while(i<n&&j<m)

    {

        if(s[i]==t[j])

        {

            i++;j++;

        }

        else

        {

            if(j==0)

                i++;

            else

            {

                j=f[j-1]+1;

            }

        }

    }

    if(j<m)return -1;

    else return i-m+1;

}

int main()

{

    int d;

    cin>>d;

    for(int j=1;j<=d;j++)

    {

        scanf("%d%d",&n,&m);

        for(int i=0;i<n;i++)

            cin>>s[i];

        for(int ii=0;ii<m;ii++)

            cin>>t[ii];

        cout<<kmp()<<endl;

    }

    return 0;

}