[HDOJ1686]Oulipo

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1686

Oulipo

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7716 Accepted Submission(s): 3102


[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
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.

[align=left]Sample Input[/align]

3
BAPC
BAPC
AZA
AZAZAZA
VERDI
AVERDXIVYERDIAN

[align=left]Sample Output[/align]

1
3
0

还是KMP，和上一个题一样。就当时熟练一下KMP。要小心的是这回匹配到最后一个字符，并且注意输入T以后要吸收掉\n。还有目标串和模式串的位置，跟上一个题也不一样。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <queue>
#include <map>
#include <stack>
#include <list>
#include <vector>

using namespace std;

const int maxn = 1000010;
int na, nb;
char a[maxn];
char b[maxn];
int pre[maxn];

void getpre(char *b, int *pre) {
int j, k;
pre[0] = -1;
j = 0;
k = -1;
while(j < nb) {
if(k == -1 || b[j] == b[k]) {//匹配
j++;
k++;
pre[j] = k;
}
else {  //b[j] != b[k]
k = pre[k];
}
}
}

int kmp() {
int ans = 0;
int i = 0;
int j = 0;
getpre(b, pre);
while(i < na) {
if(j == -1 || a[i] == b[j]) {
i++;
j++;
}
else {
j = pre[j];
}
if(j == nb) {
ans++;
}
}
return ans;
}

int main() {
int T;
scanf("%d", &T);
getchar();
while(T--) {
memset(pre, 0, sizeof(pre));
gets(b);
gets(a);
na = strlen(a);
nb = strlen(b);
printf("%d\n", kmp());
}
return 0;
}