1711 Number Sequence(kmp)

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

 

 

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题解:kmp模板 编译错误了多次,无爱啦...

 

代码:

#include <iostream>
#include <cstdio>
using namespace std;
int p[10010],s[1000010];
int n,m;
int nex[10010];

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

int kmp()
{
    int i=0,j=0;
    int slen=n;
    int plen=m;
    while(i < slen && j< plen){
        if(j==-1 || s[i]==p[j]){
            ++i;
            ++j;
        }
        else{
            j=nex[j];
        }
    }
    if(j == plen)
        return i-j+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",&s[i]);
        for(int i=0; i<m; i++)
            scanf("%d",&p[i]);
        get();
        printf("%d\n",kmp());
    }
}