hdu 1686 比较 hdu 2087 深入了解KMP

Oulipo

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 4002    Accepted Submission(s): 1579


Problem 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

 

Source

华东区大学生程序设计邀请赛_热身赛

 

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当我在比赛时看到这个题目时我嘿嘿了，因为我真的会KMP,但是随即我又呵呵了，因为我确实想不到对于这个应该是怎么做的，这让我想起来以前做过的hdu 2087的剪

华布条的题目，非常的类似，但是hdu 2087的问题在于我们对于这个字符串是不可以

重复利用的，但是求next数组的过程都是一致的，关键在于其比较。其实核心在于J

什么时候被赋成0的问题。对于这个问题来说，就是比较何时从小字符串的开始合适比。对于hdu 1686来说，j=0只能是当前面j==-1也就是没得匹配的时候才进行从0开始，可是对于剪花布条而言，因为不能重复利用，只能是等到当我们的小字符串已经被完全匹配过，sum++之后才可以进行的。

hdu 2087 点击打开链接的博客。

#include<iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
int i,j,k,t;
char s1[10010],s2[1000010];
int next[10010];
/*

当发生失配的情况下，
j的新值next[j]取决于模式串
中T[0 ~ j-1]中前缀和后缀相等部分的长度，
并且next[j]恰好等于这个最大长度。
*/
void  getNext(int len)
{
int j,k;
next[0]=-1;
j=0;
k=-1;
while(j<len)
{
if(k==-1||s1[k]==s1[j])
{
k++;
j++;
next[j]=k;
}
else
k=next[k];
}
}

int main()
{
/*
while(scanf("%s",s1))
{
if(s1[0]=='#')
break;
scanf("%s",s2);
la=strlen(s1);
lb=strlen(s2);
getNext();
int sum=0;
if(la<lb)
printf("0\n");
else
{

int i=0;
int j=0;
while(i<la)
{
if(j==-1||s1[i]==s2[j])
{
i++;
j++;
}
else
j=next[j];
if(j==lb)
{
j=0;
sum++;
}
}
printf("%d\n",sum);
}
}
return 0;

*/
scanf("%d",&t);
while(t--)
{
scanf("%s",s1);
scanf("%s",s2);
int i,j,k;
i=0;
j=0;
k=0;
int la=strlen(s2);
int lb=strlen(s1);
getNext(lb);
int sum=0;
/* for(i=0;i<=lb;i++)
{
printf("h%d\n",next[i]);
}*/
/*for(i=0;i<=lb;i++)
{
printf("h %d\n",next[i]);
}*/
while(i<la)
{
//printf("jj  %d\n",j);
if(s2[i]==s1[j])
{

i++;
j++;
//  printf("jjjj  %d\n",j);
}
else
{
//printf("%c hh  %c\n",s2[i],s1[j]);
j=next[j];
if(j==-1)
{
i++;
j=0;
}
}
if(j==lb)
{
sum++;
}

}
printf("%d\n",sum);

}
return  0;
}