Unit29 Do you have these shoes in a size nine?

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和UVa11077的分析很类似. 我们固定左脚的鞋子不动,然后将右脚的鞋子看做一个置换分解. 对于一个长度为l的循环节,要交换到正确位置至少要交换l-1次. 1 #include <cstdio> 2 #include <cstring> 3 #include <map> 4 using namespace std; 5 6 bool vis[10000 + 10]; 7 8 int main() 9 { 10 //freopen("in.txt",

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最近看到一个模型用到了shoes,准备自己试一试.搞了半天,也安装不成功.直接安装包,gem install shoes,失败, 从rubygems上下载最新版本的shoes 然后安装,成功了.随便写了一个代码, 1 require 'shoes' 2 Shoes.app do 3 button("Click me!") do 4 alert("hello shoes!") 5 end 6 end 根本不出现UI 再看一下这个包的描述,它就是没有UI的. 再看pre

题目链接:uva 11330 - Andy's Shoes 题目大意:小andy有很多鞋,穿完到处丢,后来他把所有鞋都放回鞋架排成一排,保证了鞋的左右交替,但是颜色混了.问说他至少移动多少次可以将鞋分类好. 解题思路:对应奇数位置为左鞋,偶数位置为右鞋,一双鞋只有一只左鞋和一只右鞋,保证不换左变鞋子,以左鞋的位置为基准换右边鞋子,对应右边鞋子的位置即为一个置换,将置换的循环分解为x个互不相干的循环,ans=n-x #include <cstdio> #include <cstring&g

题目:有双配对出错鞋子,要求最少的交换次数,使得鞋子配对摆放. 分析:组合数学,置换群.统计置换中循环的个数k,则结果为n-k. 循环内部(设有m个元素)需要交换m-1次(除最后一次,每次交换最多只能有一个复位) 说明:注意鞋子的编号不一定是1~n,是1~10000之间的数字,计算时需要做映射. #include <cstdlib> #include <cstdio> int value[10011]; int Lshoes[10011]; int Rshoes[10011]; i

UVA 11330 - Andy's Shoes 题目链接 题意:andy有很多双鞋子,每双鞋子有一个编号,现在他把鞋子左右左右放回去,可是不能保证所有鞋子左边和右边是同一编号,现在要求用最少的交换次数,使得所有鞋子左右编号相同 思路:置换的分解,固定左边的鞋子,这样右边的鞋子就可以看成是放在哪个位置,然后根据这个求出每个循环的长度,最后每个循环长度-1的总和就是答案 代码: #include <cstdio> #include <cstring> int t, n, vis[10

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