渣渣ACM日记——3461-Oulipo(POJ)

Oulipo

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 28717		Accepted: 11472

Description

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
0

//第一次学KMP  超时了好多次  学习学习！

AC代码：

#include <iostream>
#include <cstdio>
#include <string.h>
using namespace std;
const int MAXN=1000000+50;
int next[MAXN];
int tlen,plen;
int sum;
int main(){
	void GetNext(char *pat);
	void kmp(char *text,char *pat);
	char text[MAXN];
	char pat[10050];
	int t;
	cin>>t;
	while(t--){
		sum=0;
	scanf("%s%s",pat,text);
	tlen=strlen(text);
	plen=strlen(pat);
	GetNext(pat);
	kmp(text,pat);
	 cout<<sum<<endl;
	}
	return 0;
}
void GetNext(char *pat){
	int j=0,k=-1;
	next[0]=-1;
	while(j<plen){
		if(k==-1||pat[j]==pat[k])
			next[++j]=++k;
		else
		   k=next[k];
	}
}
void kmp(char *text,char *pat){
	int i=0,j=0;
	while(i<tlen&&j<plen){
		if(j==-1||text[i]==pat[j]){
			i++;
			j++;
		}
		else
		   j=next[j];	
	if(j==plen)
	{
	  sum++;
	  j=next[j];
}
}
}
//网上查了都说sunday比较吊 这个很多人都超时  只能用KMP  或者是我们写不对  

//超时的sunday

#include <iostream>
#include <cstdio>
#include <string.h>
using namespace std;
int book[30];
int plen,tlen;
const int MAXN=1000000+20;
int sum;
int main(){	 
	void getbook(char *pat);
	void judge(char *text,char *pat);
	char text[MAXN],pat[10005];
	int i,t;
	cin>>t;
	while(t--){
		sum=0;
	    scanf("%s%s",pat,text); 	
		plen=strlen(pat),tlen=strlen(text);
		memset(book,-1,sizeof(book));
		getbook(pat);
		judge(text,pat);		
		cout<<sum<<endl;		
	}
	return 0;
}
void getbook(char *pat){
	 int i;
	 for(i=plen-1;i>=0;i--)
	 	if(book[pat[i]-'A']==-1)
	 		book[pat[i]-'A']=i;	
    /*for(i=1;i<128;i++)
    	cout<<i<<"  "<<book[i]<<endl;
	*/	 
}
void judge(char *text,char *pat){
	int i,j,now; //j匹配了几个 
	for(i=0,j=0;i<tlen;){
		if(text[i]!=pat[j])   //不匹配 
		{
			 now=i-j+plen;   //子串后面位置对应的母串的位置 
			if(now<tlen){   //没越界 
				if(book[text[now]]==-1)//子串后面位置对应的母串没在子串出现 
					i=now+1;								
				else				
					i=now-book[text[now]-'A']; 
					//校准下次开始匹配的母串位置 
				j=0;  //下次匹配从子串首字符开始 
	    	}
			else  //越界 找不到 
	       return ;
		}		
		else        //匹配一个 
		 {
		 	i++,j++;
		 }
		if(j==plen){
		    sum++;
			j=0;  
			if(i<tlen){   
				if(book[text[i]]==-1)
					i++;								
				else				
					i=i-book[text[i]-'A']; 					
	    	}
			else 
	       return ;		    
		}	
	}
}