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:

 1 #include <iostream>

 2 #include <cstdio>

 3 using namespace std;

 4 

 5 int n,m;

 6 int a[1000010],b[10010],next[10010];

 7 

 8 void getnext (int *s,int *next){

 9     next[0]=next[1]=0;

10     for (int i=1;i<m;i++){

11         int j=next[i];

12         while (j&&s[i]!=s[j])

13             j=next[j];

14         if(s[i]==s[j]) next[i+1]=j+1;

15         else next[i+1]=0;

16     }

17 }

18 

19 int kmp (int *a,int *b,int *next){

20     getnext (b,next);

21     int j=0;

22     for (int i=0;i<n;i++){

23         while (j&&a[i]!=b[j])

24             j=next[j];

25         if (a[i]==b[j])

26             j++;

27         if (j==m)

28             return i-m+2;

29     }

30     return -1;

31 }

32 

33 int main (){

34     int t;

35     scanf ("%d",&t);

36     while (t--){

37         scanf ("%d %d",&n,&m);

38         for (int i=0;i<n;i++)

39             scanf ("%d",&a[i]);

40         for (int i=0;i<m;i++)

41             scanf ("%d",&b[i]);

42         int ans=kmp (a,b,next);

43         printf ("%d\n",ans);

44     }

45     return 0;

46 }