hdu 1711 Number Sequence

Number Sequence

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7731    Accepted Submission(s): 3520

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

想不明白，利用G++提交超时，而利用c++确accpet

#include <iostream>
#include <algorithm>
#include <vector>
#include <cstring>
#include <cstdlib>

using namespace std;

const int MAX = 10000+2;
int next[MAX] = {0};

void getNext(const vector<int> &b){
    memset(next,0,sizeof(next));
    int j = 0;
    next[1] = 0;
    for(int i = 2; i < b.size(); i ++){
        while( j > 0 && b[j+1] != b[i] ) j = next[j];
        if( b[j+1] == b[i]) j++;
        next[i] = j;
    }
}

int kmp(const vector<int> &a, const vector<int> &b){
    int j = 0;
    for(int i = 1; i < a.size(); i ++ ){
        while( j > 0 && a[i] != b[j+1]) j=next[j];
        if( a[i] == b[j+1]) j++;
        if( j == b.size()-1 ) return i-b.size()+2;
    }
    return -1;
}

int main(){
    int T;
    cin  >> T;
    while(T--){
        int n,m;
        cin >> n >> m ;
        vector <int> a(n+1),b(m+1);
        for(int i = 1; i <= n ; i ++ ) cin >> a[i];
        for(int i = 1; i <= m ; i ++ ) cin >> b[i];
        if( n < m) cout<<-1<<endl;
        else{
            getNext(b);
            cout<<kmp(a,b)<<endl;
        }
    }
    return 0;
}