本文主要是介绍UVa 188 - Perfect Hash，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
题目链接：
http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=24&page=show_problem&problem=124
类型： 哈希
原题：
Perfect Software, Inc. has obtained a government contract to examine text flowing through a high-speed network for the occurrence of certain words. Your boss, Wally Perfect, has designed a parallel processing system which checks each word against a group of small perfect hash tables.
A perfect hash function maps its input directly to a fully occupied table. Your job is to construct the perfect hash functions from the lists of words in each table. The hash function is of the form
, where C is a positive integer you are to discover, w is an integer representation of an input word, and n is the length of the table. C must be as small as possible. Note that 
is the floor function and that 
for some real number R is the largest integer that is 
.
Here are Wally's notes on the subject:
Let 
consist of positive integers 
. The problem is to find the smallest positive integer C such that
for all 
.
C must be a multiple of at least one element of W.
If some
for all 
,
then the next largest C that could resolve the conflict is at least
Since all such conflicts must be resolved, it is advantageous to choose the largest candidate from among the conflicts as the next C to test.
You are to convert each word to a number by processing each letter from left to right. Consider `a' to be 1, `b' to be 2, 
, `z' to be 26. Use 5 bits for each letter (shift left by 5 or multiply by 32). Thus `a' = 1, `bz' = 
.
Input
Input to your program will be a series of word lists, one per line, terminated by the end-of-file. Each line consists of between two and thirteen words of at most five lower case letters each, separated from each other by at least one blank. There will always be at least one one-letter word.
For each list, you are to print the input line. On the next line, print the C for the hash function determined by the list. Print a blank line after each C.
C will always fit in a 32-bit integer.
Sample input
this is a test of some words to try out a bee see dee the of and to a in that is i it with for as
Sample output
this is a test of some words to try out 17247663a bee see dee 4427the of and to a in that is i it with for as 667241
题目大意 + 分析与总结：
真心不喜欢这种题目， 题意很难弄懂，而弄懂了之后就像切菜一样地切。
有一个完美哈希函数，其中，C是一个正数， 也就是你要找的那个数（结果要输出这个数）。w是由一个单词转换得到的数字，例如 `a' = 1, `bz' = ， 可以把它看成是32进制的转换。 n其实就是代表的是一行中的单词的个数。
然后怎样求出C呢？
首先对于， .是由一行中的各个单词转换而来的，然后题目说C must be a multiple of at least one element of W. 也就是C必须是W中某一个的倍数， 然后再上面点还说C must be as small as possible. C必须尽可能地小。 所以，在开始时， 让C等于w1（w1是最小的，因为W集合已经排好序了：）。
对于C，要让它符合条件 for all . 所以要用一个两层for循环来判断。
如果不符合的话，就让C等于：
一直到找到符合的条件C为止，答案就出来了。
代码：
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n, W[15], C, ans; inline int min(int a, int b){ return a<b?a:b;
}
// 递归找出符合条件的C
void solve( ){ for(int i=0; i<n; ++i){ for(int j=i+1; j<n; ++j)if((C/W[i])%n==(C/W[j])%n){ C = min((C/W[i]+1)*W[i], (C/W[j]+1)*W[j]); solve(); return ; } }
} int main(){ char str[200]; int i; while(gets(str)){ n=0; memset(W, 0, sizeof(W)); for(i=0; i<=strlen(str); ++i){ if(str[i]==' ' || str[i]=='\0'){ ++n; while(str[i+1]==' ')++i; // 跳过连续的空格} else{ W[n] = (W[n]<<5)+str[i]-'a'+1; } } sort(W, W+n); // 排序，让w1<w2<w3...<wnC = W[0];solve(); puts(str); printf("%d\n", C); puts(""); } return 0;
}
—— 生命的意义，在于赋予它意义。
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