HDU 1711 Number Sequence (KMP找子串第一次出现的位置）

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5898    Accepted Submission(s): 2652

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>

int a[1000010],b[10010];

int next[10010];

int n,m;

void getNext()

{

    int j,k;

    j=0;

    k=-1;

    next[0]=-1;

    while(j<m)

    {

        if(k==-1||b[j]==b[k])

          next[++j]=++k;

        else k=next[k];

    }    

}  

//返回首次出现的位置 

int KMP_Index()

{

    int i=0,j=0;

    getNext();

    

    while(i<n && j<m)

    {

        if(j==-1||a[i]==b[j])

        {

            i++;

            j++;

        }    

        else j=next[j];

        

    }    

    if(j==m) return i-m+1;

    else return -1;

}      

int main()

{

    int T;

    scanf("%d",&T);

    while(T--)

    {

        scanf("%d%d",&n,&m);

        for(int i=0;i<n;i++)

          scanf("%d",&a[i]);

        for(int i=0;i<m;i++)

          scanf("%d",&b[i]);

        printf("%d\n",KMP_Index());

    }    

    return 0;

}    

 

#include<stdio.h>

#include<iostream>

#include<string.h>

#include<algorithm>

using namespace std;

const int MAXN=1000010;



int a[MAXN];

int b[MAXN];



int n,m;

int next[MAXN];

/*

根据定义next[0]=-1，假设next[j]=k, 即P[0...k-1]==P[j-k,j-1]

1)若P[j]==P[k]，则有P[0..k]==P[j-k+1,j]，很显然，next[j+1]=next[j]+1=k+1;

2)若P[j]!=P[k]，则可以把其看做模式匹配的问题，即匹配失败的时候，k值如何移动，显然k=next[k]。

*/



void getNext()

{

    int j,k;

    next[0]=-1;

    j=0;

    k=-1;

    while(j<m)

    {

        if(k==-1||b[j]==b[k])

        {

            j++;

            k++;

            next[j]=k;

        }

        else k=next[k];

    }

}



int KMP_Index()

{

    int i=0,j=0;

    getNext();

    while(i<n&&j<m)

    {

        if(j==-1||a[i]==b[j])

        {

            i++;j++;

        }

        else j=next[j];

    }

    if(j==m)return i-m+1;

    else return -1;

}

int main()

{

    int T;

    scanf("%d",&T);

    while(T--)

    {

        scanf("%d%d",&n,&m);

        for(int i=0;i<n;i++)scanf("%d",&a[i]);

        for(int i=0;i<m;i++)scanf("%d",&b[i]);

        printf("%d\n",KMP_Index());

    }

    return 0;

}