【kmp】POJ-3461 Oulipo

附上题目链接：http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=10368

Oulipo

Time Limit: 1000MS

Memory Limit: 65536KB

64bit IO Format: %I64d & %I64u

Submit Status

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

BAPC 2006 Qualification

不多说，一道裸的kmp。。。只不过是求个数而不是求位置，于是乎刚开始时被光荣的坑了。。。找到一个后不是j=0；而是j = next[j]，因为要充分利用好next这个资源，next是最大前缀后缀相同的长度。。。。也就是说w串只需移动j - next[j]个长度就够了，因为前面是相同的，不需重新比较。。

【代码】：

#include<cstdio>
#include<iostream>
#include<cstring>
using namespace std;

char w[10000 + 5];
char t[1000000 + 5];
int next[10000 + 5];
int T;
int ans = 0;

void getnext(int &w_len)
{
int k = -1,j = 0;
next[0] = -1;
while(j < w_len)
{
if(k == -1 || w[k] == w[j])
{
j++;k++;
if(w[j] != w[k]) next[j] = k;
else next[j] = next[k];
}
else k = next[k];
}
}
void kmp(int &w_len,int &t_len)
{
int i = 0, j = 0;
while(i < t_len && j < w_len)
{
if(t[i] == w[j] || j == -1)
{
i++;j++;
}
else j = next[j];
if(j == w_len){
ans++;
j = next[j];
}
}
}
int main()
{
scanf("%d", &T);
while(T--)
{
ans = 0;
scanf("%s",w);
scanf("%s",t);
int w_len = strlen(w);
int t_len = strlen(t);
getnext(w_len);
kmp(w_len, t_len);
printf("%d\n",ans);
}
return 0;
}