http://acm.hdu.edu.cn/showproblem.php?pid=1533 Going Home Problem Description   On a grid map there are n little men and n houses. In each unit time, every little man can move one unit step, either horizontally, or vertically, to an adjacent point. F…

Going Home Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 2963    Accepted Submission(s): 1492 Problem Description On a grid map there are n little men and n houses. In each unit time, every…

The Windy's Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 4158   Accepted: 1777 Description The Windy's is a world famous toy factory that owns M top-class workshop to make toys. This year the manager receivesN orders for toys. The man…

带权二分图匹配,把距离当做权值,因为是最小匹配,所以把距离的相反数当做权值求最大匹配. 最后再把答案取一下反即可. #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <map> #include <vector> #include <cmath> #define MP make_pair using…

传送门:http://poj.org/problem?id=2195 Going Home Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 26151   Accepted: 13117 Description On a grid map there are n little men and n houses. In each unit time, every little man can move one unit st…

题意: 给出一个有向带权图,找到若干个圈,使得每个点恰好属于一个圈.而且这些圈所有边的权值之和最小. 分析: 每个点恰好属于一个有向圈 就等价于 每个点都有唯一后继. 所以把每个点i拆成两个点,Xi 和 Yi ,然后求二分图最小权完美匹配(流量为n也就是满载时,就是完美匹配). #include <bits/stdc++.h> using namespace std; + ; ; struct Edge { int from, to, cap, flow, cost; Edge(int u,…

KM算法大概过程: (1)初始化Lx数组为该boy的一条权值最大的出边.初始化Ly数组为 0. (2)对于每个boy,用DFS为其找到一个girl对象,顺路记录下S和T集,并更新每个girl的slack值.若不能为其找到对象,则转3. (3)找出非T集合的girl的最小slack值为d,更新S集中的boy和T集中的girl,并且顺路更新非T集中的slack值.对于那个失败的boy继续第2步. 简括之,就是在保持当前权和最高的情况下,尽量为每个boy找到权更大的边.找的过程就是DFS过程,标记出S…

由于涉及到实数,一定,一定不能直接等于,一定,一定加一个误差<0.00001,坑死了…… 有两种事物,不难想到用二分图.这里涉及到一个有趣的问题,这个二分图的完美匹配的最小权值和就是答案.为啥呢?因为如果有四个点,a,b,c,d .Ab和cd交叉,ac和bd不交叉,那么ac和bd的长度和一定小于ab和cd的长度和,可以画一个图很容易就证出来.所以,如果所有的边都不交叉,又因为有解,那么最小的权值和就是解了.附图一枚,自己画的,比较简陋,凑活着看吧…… 用KM算法求最佳完美匹配最小权值和,可以直接…

[POJ 2195] Going Home(KM算法求最小权匹配) Going Home Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 20303   Accepted: 10297 Description On a grid map there are n little men and n houses. In each unit time, every little man can move one unit ste…

/** 题目:poj3565 Ants km算法求最小权完美匹配,浮点权值. 链接:http://poj.org/problem?id=3565 题意:给定n个白点的二维坐标,n个黑点的二维坐标. 求是否存在n条边,每条边恰好连一个白点,一个黑点,且所有的边不相交. 输出所有黑点连接的白点编号. 思路:最小权完美匹配. 假定有白点1(a1,b1), 2(a2,b2), 黑点3(a3,b3),4(a4,b4); 如果1(a1,b1)与3(a3,b3)相连,2(a2,b2)与4(a4,b4)相连,如…