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[],b[],next[];

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

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

 int main (){
     int t;
     scanf ("%d",&t);
     while (t--){
         scanf ("%d %d",&n,&m);
         for (int i=;i<n;i++)
             scanf ("%d",&a[i]);
         for (int i=;i<m;i++)
             scanf ("%d",&b[i]);
         int ans=kmp (a,b,next);
         printf ("%d\n",ans);
     }
     return ;
 }