Number Sequence

Number Sequence

Time Limit : 10000/5000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other)

Total Submission(s) : 22   Accepted Submission(s) : 12

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

先输入T，表示有T组测试样例。
然后输入两个数据，表示两个有多长的字符长度，然后匹配两个字符串，返回能够匹配第一点的位置

 

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int S[1111110],t[200000],Len_S,Len_t;
void getNext(int p[],int next[])
{
    int j,k,L;
    next[0]=-1;
    j=0;
    k=-1;
    while(j<Len_t-1)
    {
        if(k==-1||p[j]==p[k])
        {
            j++;
            k++;
            next[j]=k;
        }
        else
            k=next[k];
    }
}

int KMPMatch(int s[],int p[])
{
    int next[200000];
    int i,j;
    i=0;
    j=0;
    getNext(p,next);
    while(i<Len_S)
    {
        if(j==-1||s[i]==p[j])
        {
            i++;
            j++;
        }
        else
        {
            j=next[j];
        }
        if(j==Len_t)
            return i-Len_t+1;
    }
    return -1;
}
int main()
{
    int T,i,Sign;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d%d",&Len_S,&Len_t);
        for(i=0;i<Len_S;i++)
            scanf("%d",&S[i]);
        for(i=0;i<Len_t;i++)
            scanf("%d",&t[i]);
        printf("%d\n",KMPMatch(S,t));/*匹配返回长度*/
    }
    return 0;
}

 KMP模版（2016.4.19）

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int Num[2][1100111];
int Len[2];
void getNext(int p[],int next[],int Len_t)
{
    int i=0,j=-1;
    next[0]=-1;
    while(i<Len_t)
    {
        if(j==-1||p[i]==p[j])next[++i]=++j;
        else j=next[j];
    }
}
int KMPMatch(int s[],int p[])
{
    int next[200005];
    int i=0,j=0,sign=0;
    int Len_t=Len[1],Len_S=Len[0];
    getNext(p,next,Len_t);
    while(i<Len_S)
    {
        if(j==-1||s[i]==p[j]){i++;j++;}
        else j=next[j];
        if(j==Len_t) return i-Len_t+1;
    }
    return -1;    /*返回值是子串在主串第一个匹配的起始位置*/
}

int main()
{
    int T;
    scanf("%d",&T);
    while(T--){
        scanf("%d%d",&Len[0],&Len[1]);
        for(int i=0;i<Len[0];i++)scanf("%d",&Num[0][i]);
        for(int i=0;i<Len[1];i++)scanf("%d",&Num[1][i]);
        printf("%d\n",KMPMatch(Num[0],Num[1]));
    }
    return 0;
}