读题显然是二分图匹配,看成guest与umbrella的匹配.匈牙利果断TLE了,其实时间卡的相当紧.HK过的,750ms. /* 2389 */ #include <iostream> #include <string> #include <map> #include <queue> #include <set> #include <stack> #include <vector> #include <algori…

HDU 2389 Rain on your Parade / HUST 1164 4 Rain on your Parade(二分图的最大匹配) Description 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 send…

Rain on your Parade Time Limit:3000MS     Memory Limit:165535KB     64bit IO Format:%I64d & %I64u Submit Status Practice HDU 2389 Description You’re giving a party in the garden of your villa by the sea. The party is a huge success, and everyone is h…

 HK.... Rain on your Parade Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 655350/165535 K (Java/Others)Total Submission(s): 2229    Accepted Submission(s): 696 Problem Description You're giving a party in the garden of your villa by the sea…

Rain on your Parade Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 655350/165535 K (Java/Others)Total Submission(s): 2154    Accepted Submission(s): 662 Problem Description You’re giving a party in the garden of your villa by the sea. The pa…

<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>…

其实是求树上的路径间的数据第K大的题目.果断主席树 + LCA.初始流量是这条路径上的最小值.若a<=b,显然直接为s->t建立pipe可以使流量最优:否则,对[0, 10**4]二分得到boundry,使得boundry * n_edge - sum_edge <= k/b, 或者建立s->t,然后不断extend s->t. /* 4729 */ #include <iostream> #include <sstream> #include <…

Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining. For example, Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6. The above elevation map is represented by a…

DP/四边形不等式 裸题环形石子合并…… 拆环为链即可 //HDOJ 3506 #include<cmath> #include<vector> #include<cstdio> #include<cstring> #include<cstdlib> #include<iostream> #include<algorithm> #define rep(i,n) for(int i=0;i<n;++i) #define…

DP/四边形不等式 这题跟石子合并有点像…… dp[i][j]为将第 i 个点开始的 j 个点合并的最小代价. 易知有 dp[i][j]=min{dp[i][j] , dp[i][k-i+1]+dp[k+1][j-(k-i+1)]+w(i,k,j)} (这个地方一开始写错了……) 即,将一棵树从k处断开成(i,k)和(k+1,i+j-1)两棵树,再加上将两棵树连起来的两条树枝的长度w(i,k,j) 其中,$ w(i,k,j)=x[k+1]-x[i]+y[k]-y[i+j-1] $ 那么根据四边形…