[POJ 3461] Oulipo【自然溢出hash】
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
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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