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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(b[i]==b[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 }
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