[POJ 3461] Oulipo【自然溢出hash】

  • 2018-01-07
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Problem:

Time Limit: 1000MS Memory Limit: 65536K

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

Solution:

一道很好的字符串 hash 题。(当然正解是 KMP 模板啦 .囧. 

KMP 解法请见: [POJ 3461] Oulipo【KMP】

刚开始没有过,大概是 hash 冲突了,改成多重 hash 之后又 TLE 掉了。。

于是乎在网上搜到了一种 看起来很不靠谱 神奇的 hash,就 莫名其妙地 水过了。。

【神奇的 hash】

似乎是 unsigned long long 的自然溢出版的。

特点是取的系数特别大,似乎可以降低冲突的概率。

但是目前还不知道普遍效果如何以及如何证明。。先用着,以后慢慢研究。。

Code: O(TL), L为文章长度 [1120K, 235MS]

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<algorithm>
#define BASE 19260817  // A very large base
using namespace std;
typedef unsigned long long ull;

int cas, ans;
char W[10005], T[1000005];
int lenW, lenT;
ull hashW, hashT, topwt;

int main(){
	scanf("%d", &cas);
	while(cas--){
		scanf("%s%s", W, T);
		lenW = strlen(W), lenT = strlen(T);
		hashW = hashT = 0, topwt = 1;
		for(register int i = 0; i < lenW; i++){
			topwt *= BASE;
			hashW = hashW * BASE + W[i] - 'A';
			hashT = hashT * BASE + T[i] - 'A';
		}  // Needn't mod and let it overflow naturally
		ans = (hashW == hashT);
		for(register int i = lenW; i < lenT; i++){
			hashT = hashT * BASE + T[i] - 'A' - topwt * (T[i - lenW] - 'A');
			ans += (hashW == hashT);
		}
		printf("%d\n", ans);
	}
	return 0;
}

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OPEN AT 2017.12.10

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