F - Oulipo(kmp经典模板题）！！！

The French author Georges Perec (1936–1982) once wrote a book, La disparition, without the letter 'e'. He was a member of the Oulipo group. A quote from the book:

Tout avait Pair normal, mais tout s’affirmait faux. Tout avait Fair normal, d’abord, puis surgissait l’inhumain, l’affolant. Il aurait voulu savoir où s’articulait l’association qui l’unissait au roman : stir son tapis, assaillant à tout instant son imagination, l’intuition d’un tabou, la vision d’un mal obscur, d’un quoi vacant, d’un non-dit : la vision, l’avision d’un oubli commandant tout, où s’abolissait la raison : tout avait l’air normal mais…

Perec would probably have scored high (or rather, low) in the following contest. People are asked to write a perhaps even meaningful text on some subject with as few occurrences of a given “word” as possible. Our task is to provide the jury with a program that counts these occurrences, in order to obtain a ranking of the competitors. These competitors often write very long texts with nonsense meaning; a sequence of 500,000 consecutive 'T's is not unusual. And they never use spaces.

So we want to quickly find out how often a word, i.e., a given string, occurs in a text. More formally: given the alphabet {'A', 'B', 'C', …, 'Z'} and two finite strings over that alphabet, a word W and a text T, count the number of occurrences of W in T. All the consecutive characters of W must exactly match consecutive characters of T. Occurrences may overlap.

Input

The first line of the input file contains a single number: the number of test cases to follow. Each test case has the following format:

One line with the word W, a string over {'A', 'B', 'C', …, 'Z'}, with 1 ≤ |W| ≤ 10,000 (here |W| denotes the length of the string W).
One line with the text T, a string over {'A', 'B', 'C', …, 'Z'}, with |W| ≤ |T| ≤ 1,000,000.

Output

For every test case in the input file, the output should contain a single number, on a single line: the number of occurrences of the word W in the text T.

Sample Input

3
BAPC
BAPC
AZA
AZAZAZA
VERDI
AVERDXIVYERDIAN

Sample Output

1
3

解题思路

kmp完全匹配。。。注意next[i]=j中保存的是：b[]中以第i个数（b[i-1]）结尾的前j个数（包括b[i-1]）匹配b[]的前j个数。

AC代码

第一种方法：简单但不太好理解

#include<stdio.h>
#include<string.h>
char s[10010],p[1000010];
int ne[10010];
void getnext(int m)
{int i=0,j=-1;ne[0]=-1;while(i<m){if(j==-1||s[i]==s[j]){i++;j++;ne[i]=j; }elsej=ne[j];}
//  for(int k=0;k<i;k++)
//  printf("%d",ne[k]);
}
int kmp(int m,int n)
{getnext(m);int i=0,j=0,sum=0;while(i<n){if(j==-1||p[i]==s[j]){i++;j++;   if(j==m){sum++;j=ne[j];}}elsej=ne[j];}return sum;
}
int main()
{int t;scanf("%d",&t);while(t--){
//      memset(ne,0,sizeof(ne));scanf("%s",s);getchar();scanf("%s",p);int m=strlen(s);int n=strlen(p);printf("%d\n",kmp(m,n));}return 0;
}

第二种方法：复杂但好理解

#include<stdio.h>
#include<iostream>
#include<string.h>
using namespace std;
char a[10010],b[1000010];
int ne[10010];
int main()
{int t;scanf("%d",&t);while(t--){scanf("%s",a);getchar();scanf("%s",b);int alen=strlen(a);int blen=strlen(b);ne[0]=-1;for(int i=1,j=-1;i<alen;i++){while(j>=0&&a[i]!=a[j+1]) j=ne[j];if(a[i]==a[j+1])j++;ne[i]=j;} int sum=0;for(int i=0,j=-1;i<blen;i++){while(j!=-1&&b[i]!=a[j+1]) {j=ne[j];}if(b[i]==a[j+1]) j++;if(j==alen-1){sum++;j=ne[j];}}printf("%d\n",sum);}return 0;
}