http://acm.hdu.edu.cn/showproblem.php?pid=4424 [题目大意] 给你N个点和N-1条边的连通图,也就是说任意两点间的路径是唯一的.每条边有个权值,从一点到另一点的最大流量是路径中所有边中权值的最小值.要你找一个点,使得这点到剩下所有点的最大流量之和最大,求这个最大流量和. [分析1:动态规划 ] 假设现在有两个点的集合S1和S2,添加的一条边可以使得这两个集合连通起来,要求最大的流量之和,那么只有两种情况:1.选取的中心点在S1中:2.选取的中心点在S…

这题要用到一点贪心的思想,因为一个点到另一个点的运载能力决定于其间的边的最小权值,所以先把线段按权值从大到小排个序,每次加的边都比以前小,然后合并集合时,比较 x = findset(a) 做根或 y = findset(b) 做根时,总权值的大小,x做根的总权值 ca = num[b]*w + cap[a] ,b同理.即b这个集合的点个数乘以新加的边的距离为新增的权值.然后合并.. 代码: #include <iostream> #include <cstdio> #includ…

///题意:给出一棵树.树的边上都有边权值,求从一点出发的权值和最大,权值为从一点出去路径上边权的最小值 # include <stdio.h> # include <algorithm> # include <iostream> # include <string.h> using namespace std; # define MAX 200010 struct node { int u,v; int w; }; struct node a[MAX];…

给你一颗树 每条边有一个权值 选择一个点为中心 定义S值为中心到其它n-1个点的路径上的最小边权 求全部点S值的和 从大到小排序 每次合并2棵树 设为A集合 B集合 设A集合的最大S值的和为sumA B集合为sumB 中心在A或者B如今增加A-B这条边使得2个集合连通 由于A-B这条边的权值小于等于AB集合里面边的权值 所以假设合并之后中心在A 那么sumA = sumA+B集合的点数*A-B这条边的权值 sumB = sumB+A集合的点数*A-B这条边的权值 2者取大 #include <c…

并检查集合 侧降序,每增加一个侧面应该推断,其中基本建设方..... Conquer a New Region Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 1169    Accepted Submission(s): 373 Problem Description The wheel of the history rolling…

Conquer a New Region Time Limit: 8000/4000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 657 Accepted Submission(s): 179 Problem Description The wheel of the history rolling forward, our king conquered a new region i…

Conquer a New Region Time Limit: 5 Seconds      Memory Limit: 32768 KB The wheel of the history rolling forward, our king conquered a new region in a distant continent. There are N towns (numbered from 1 to N) in this region connected by several road…

Conquer a New Region Time Limit: 5 Seconds      Memory Limit: 32768 KB The wheel of the history rolling forward, our king conquered a new region in a distant continent. There are N towns (numbered from 1 to N) in this region connected by several road…

The wheel of the history rolling forward, our king conquered a new region in a distant continent.There are N towns (numbered from 1 to N) in this region connected by several roads. It's confirmed that there is exact one route between any two towns. T…

思路:将边从大到小排序,判断向哪边连,能使总和最大. #include<map> #include<set> #include<cmath> #include<queue> #include<cstdio> #include<vector> #include<string> #include<cstdlib> #include<cstring> #include<iostream> #i…