HDU1711 最基础的kmp算法

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.

 

 

这是一道对kmp算法的最基础运用，在这里主要注意kmp函数以及得到next的函数的写法

 

代码如下:

#include <iostream>
#include <cstdio>
using namespace std;

int n,m;
int a[1000010],b[10010],next[10010];

void getnext (int *s,int *next){
    next[0]=next[1]=0;
    for (int i=1;i<m;i++){
        int j=next[i];
        while (j&&s[i]!=s[j])
            j=next[j];
        if(s[i]==s[j]) next[i+1]=j+1;
        else next[i+1]=0;
    }
}

int kmp (int *a,int *b,int *next){
    getnext (b,next);
    int j=0;
    for (int i=0;i<n;i++){
        while (j&&a[i]!=b[j])
            j=next[j];
        if (a[i]==b[j])
            j++;
        if (j==m)
            return i-m+2;
    }
    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]);
        int ans=kmp (a,b,next);
        printf ("%d
",ans);
    }
    return 0;
}