「POJ-3461」Oulipo-KMP

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 \cdots

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’, \cdots , ‘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.

链接

POJ-3461

题解

裸题,KMP直接上模板…

代码

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#include <iostream>
#include <cstring>
#include <vector>
#include <algorithm>
#include <cmath>
#include <cstdio>
const int MAXN = 1000010;
inline void getNext(int *next, const char *p, const int len) {
for (register int i = 1, j = 0; i < len; i++) {
while (j && p[i] != p[j]) j = next[j - 1];
if (p[i] == p[j]) j++;
next[i] = j;
}
}
inline void kmp(const char *p, const char *t) {
register int n = strlen(t), m = strlen(p);
static int next[MAXN];
register int ans = 0;
memset(next, 0, sizeof(next));
getNext(next, p, m);
for (register int i = 0, j = 0; i < n; i++) {
while (j && t[i] != p[j]) j = next[j - 1];
if (t[i] == p[j]) j++;
if (j == m) ans++;
}
std::cout << ans << "\n";
}
char t[MAXN], p[MAXN];
int T;
int main() {
#ifndef ONLINE_JUDGE
freopen("in.in" , "r", stdin);
#endif
std::ios::sync_with_stdio(0);
std::cin.tie(0);
std::cin >> T;
while (T--) {
std::cin >> t >> p;
kmp(t, p);
}
return 0;
}
#

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