Ladies' Choice Time Limit: 6000ms Memory Limit: 131072KB This problem will be judged on UVALive. Original ID: 398964-bit integer IO format: %lld      Java class name: Main Background Teenagers from the local high school have asked you to help them wi…

题意:男女各n人,进行婚配,对于每个人来说,所有异性都存在优先次序,即最喜欢某人,其次喜欢某人...输出一个稳定婚配方案.所谓稳定,就是指未结婚的一对异性,彼此喜欢对方的程度都胜过自己的另一半,那么这对异性就有私奔的可能性. 这里明显也是一个匹配问题,与最佳二分完美匹配所不同的是,二分匹配重在求最佳权值,而这里求的是满足彼此的要求.这里用到了Gale-Shapley算法. 算法学习:http://wenku.baidu.com/view/964a503843323968011c92c3.html…

经典的稳定婚姻匹配问题 UVALive - 3989 Ladies' Choice Time Limit: 6000MS Memory Limit: Unknown 64bit IO Format: %lld & %llu [Submit]   [Go Back]   [Status] Description Problem I – LadiesÕ Choice Background Teenagers from the local high school have asked you to h…

https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1990 题目大意: 在盛大的校园舞会上有n位男生和n位女生,每人都对每个异性有一个排序,代表对他们的喜欢程度.你的任务是将男生和女生一一配对,使得男生U和女生V不存在一下情况1.男生u和女生v不是舞伴2,他们喜欢对方的程度都大于各自当前舞伴的程度.如果出现了2中的情况,他们可能…

/** 题目: Ladies' Choice UVALive - 3989 链接:https://vjudge.net/problem/UVALive-3989 题意:稳定婚姻问题 思路: gale_shapley算法,参考文档:https://wenku.baidu.com/view/7aa841f2fab069dc502201cb.html */ #include <iostream> #include <cstring> #include <cstdio> #in…

Ladies' Choice Teenagers from the local high school have asked you to help them with the organization of next year'sProm. The idea is to find a suitable date for everyone in the class in a fair and civilized way. So,they have organized a web site whe…

ayout: post title: 训练指南 UVALive - 3989(稳定婚姻问题) author: "luowentaoaa" catalog: true mathjax: true tags: - 二分图匹配 - 图论 - 训练指南 Ladies' Choice UVALive - 3989 #include<bits/stdc++.h> using namespace std; typedef long long ll; const ll mod=998244…

题目链接: 题目 Ladies' Choice Time Limit: 6000MS Memory Limit: Unknown 64bit IO Format: %lld & %llu 问题描述 Teenagers from the local high school have asked you to help them with the organization of next year's Prom. The idea is to find a suitable date for eve…

1175 - Ladies' Choice 链接 稳定婚姻问题. 代码: #include<bits/stdc++.h> using namespace std; typedef long long LL; inline int read() { ,f=;; +ch-';return x*f; } ; int pref[N][N],order[N][N],Boys[N],Girls[N],cur[N]; queue<int>q; // 求婚队列 void Link(int u,in…

题目大意:稳定婚姻问题.... 题目分析:模板题. 代码如下: # include<iostream> # include<cstdio> # include<queue> # include<cstring> # include<algorithm> using namespace std; # define LL long long # define REP(i,s,n) for(int i=s;i<n;++i) # define CL…