题目链接:https://vjudge.net/problem/HDU-2389 Rain on your Parade Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 655350/165535 K (Java/Others)Total Submission(s): 4889    Accepted Submission(s): 1612 Problem Description You’re giving a party in t…

Rain on your Parade Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 655350/165535 K (Java/Others)Total Submission(s): 6752    Accepted Submission(s): 2117 Problem Description You’re giving a party in the garden of your villa by the sea. The p…

You’re giving a party in the garden of your villa by the sea. The party is a huge success, and everyone is here. It’s a warm, sunny evening, and a soothing wind sends fresh, salty air from the sea. The evening is progressing just as you had imagined.…

Rain on your Parade Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 655350/165535 K (Java/Others)Total Submission(s): 5755    Accepted Submission(s): 1900 Description: You’re giving a party in the garden of your villa by the sea. The party is…

这个题唯一需要说的就是普通的匈牙利算法是O(nm)的,过不了 然后HK算法可以O(n^0.5m),这个算法可以每次找很多同样长度的最短增广路 分析见:http://www.hardbird.net/lightoj-1356-prime-independence%E6%9C%80%E5%A4%A7%E7%8B%AC%E7%AB%8B%E9%9B%86-hopcroft-carp%E7%AE%97%E6%B3%95/ #include <cstdio> #include <iostream&…

<题目链接> 题目大意:有m个宾客,n把雨伞,预计时间t后将会下大雨,告诉你每个宾客的位置和速度,每把雨伞的位置,问你最多几个宾客能够拿到伞. 解题分析: 本题就是要我们求人与伞之间的最大匹配,但是数据量较大,匈牙利算法复杂度为$O(n*m)$,会超时,所以这里用的是复杂度为$O(sqrt(n)*m)$的HK算法. 当然,本题也可以用Dinic最大流来求解最大匹配,并且这种方法求得到最大匹配的时间复杂度也为$O(sqrt(n)*m)$ #include <bits/stdc++.h>…

Rain on your Parade Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 655350/165535 K (Java/Others)Total Submission(s): 3675    Accepted Submission(s): 1187 Problem Description You’re giving a party in the garden of your villa by the sea. The p…

COURSES Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 17777   Accepted: 7007 Description Consider a group of N students and P courses. Each student visits zero, one or more than one courses. Your task is to determine whether it is poss…

1.其实HK算法思想很朴实,就是在最小均方误差准则下求得权矢量. 他相对于感知器算法的优点在于,他适用于线性可分和非线性可分得情况,对于线性可分的情况,给出最优权矢量,对于非线性可分得情况,能够判别出来,以退出迭代过程. 2.在程序编制过程中,我所受的最大困扰是:关于收敛条件的判决. 对于误差矢量:e=x*w-b 若e>0 则继续迭代 若e=0 则停止迭代,得到权矢量 若e〈0 则停止迭代,样本是非线性可分得, 若e有的分量大于0,有的分量小于0 ,则在各分量都变成零,或者停止由负值转变成正值时…

传送门:Rain on your Parade 题意:t个单位时间后开始下雨,给你N个访客的位置(一维坐标平面内)和他的移动速度,再给M个雨伞的位置,问在下雨前最多有多少人能够拿到雨伞(两个人不共用一把伞). 分析:这题匈牙利算法撸不过,只好去学习Hopcroft-Carp算法,复杂度为O(sqrt(V)*E),该算法预先找好增广路径集,避免了许多没不要的匹配. #include <cstdio> #include <cstring> #include <string>…