hdu 1711Number Sequence (KMP——输出模式串第一次出现位置)

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 15052    Accepted Submission(s): 6597

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

 

KMP模板题，返回模式串第一次出现在主串中的位置

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<cstdio>
#include<string>
#include<iostream>
#include<cstring>
#include<cmath>
#include<stack>
#include<queue>
#include<vector>
#include<map>
#include<stdlib.h>
#include<algorithm>
#define LL __int64
using namespace std;
const int MAXN=1000000+5;
int S[MAXN],T[MAXN];
int nex[MAXN];
int n,m;

void KMP_nex()
{
    int j,k;
    j=0;k=-1,nex[0]=-1;
    while(j<m)
    {
        if(k==-1 || T[j]==T[k])
            nex[++j]=++k;
        else k=nex[k];
    }
}

int KMP_ID()
{
    int i=0,j=0;
    KMP_nex();

    while(i<n && j<m)
    {
        if(j==-1 || S[i]==T[j])
            i++,j++;
        else
            j=nex[j];
    }
    if(j==m) return i-m+1;
    else return -1;
}

int main()
{
    //freopen("in.txt","r",stdin);
    int kase;
    scanf("%d",&kase);
    while(kase--)
    {
        scanf("%d %d",&n,&m);
        for(int i=0;i<n;i++) scanf("%d",&S[i]);
        for(int i=0;i<m;i++) scanf("%d",&T[i]);
        printf("%d\n",KMP_ID());
    }
    return 0;
}