Minimal Ratio Tree Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) [Problem Description] For a tree, which nodes and edges are all weighted, the ratio of it is calculated according to the following equation.Given a…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=2489 Problem Description For a tree, which nodes and edges are all weighted, the ratio of it is calculated according to the following equation. Given a complete graph of n nodes with all nodes and edges…

Minimal Ratio Tree Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 32768/32768K (Java/Other) Total Submission(s) : 12   Accepted Submission(s) : 7 Font: Times New Roman | Verdana | Georgia Font Size: ← → Problem Description For a tree, which n…

Description For a tree, which nodes and edges are all weighted, the ratio of it is calculated according to the following equation. Given a complete graph of n nodes with all nodes and edges weighted, your task is to find a tree, which is a sub-graph…

Minimal Ratio Tree Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3345    Accepted Submission(s): 1019 Problem Description For a tree, which nodes and edges are all weighted, the ratio of it i…

想到枚举m个点,然后求最小生成树,ratio即为最小生成树的边权/总的点权.但是怎么枚举这m个点,实在不会.网上查了一下大牛们的解法,用dfs枚举,没想到dfs还有这么个作用. 参考链接:http://blog.csdn.net/xingyeyongheng/article/details/9373271 #include <stdio.h> #include <string.h> #include <set> #include <vector> #incl…

http://acm.hdu.edu.cn/showproblem.php?pid=2489 这道题就是n个点中选择m个点形成一个生成树使得生成树的ratio最小.暴力枚举+最小生成树. #include <cstdio> #include <cstring> #include <algorithm> #define maxn 1000 using namespace std; <<; ; int map[maxn][maxn]; int g[maxn][m…

Minimal Ratio Tree Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 2382    Accepted Submission(s): 709 Problem Description For a tree, which nodes and edges are all weighted, the ratio of it is…

题意:给定完全无向图,求其中m个子节点,要求Sum(edge)/Sum(node)最小. 思路:由于N很小,枚举所有可能的子节点可能情况,然后求MST,memset()在POJ G++里面需要cstring头文件. #include <iostream> #include <vector> #include <map> #include <cmath> #include <memory> #include <algorithm> #i…

Minimal Ratio Tree HDU - 2489 暴力枚举点,然后跑最小生成树得到这些点时的最小边权之和. 由于枚举的时候本来就是按照字典序的,不需要额外判. 错误原因:要求输出的结尾不能有空格. #include<cstdio> #include<cstring> #include<vector> using namespace std; ],ok2[]; ]; int num,n,m; ],b[][]; vector<int> vec; dou…