E - Minimum Spanning Tree Gym - 102220E (转化+贡献)
In the mathematical discipline of graph theory, the line graph of a simple undirected weighted graph G is another simple undirected weighted graph L(G) that represents the adjacency between every two edges in G
.
Precisely speaking, for an undirected weighted graph G
without loops or multiple edges, its line graph L(G)
is a graph such that:
- Each vertex of L(G)
represents an edge of G
- .
- Two vertices of L(G)
are adjacent if and only if their corresponding edges share a common endpoint in G
- , and the weight of such edge between this two vertices is the sum of their corresponding edges' weight.
A minimum spanning tree(MST) or minimum weight spanning tree is a subset of the edges of a connected, edge-weighted undirected graph that connects all the vertices together, without any cycles and with the minimum possible total edge weight. That is, it is a spanning tree whose sum of edge weights is as small as possible.
Given a tree G
, please write a program to find the minimum spanning tree of L(G)
.
Input
The first line of the input contains an integer T(1≤T≤1000)
, denoting the number of test cases.
In each test case, there is one integer n(2≤n≤100000)
in the first line, denoting the number of vertices of G
.
For the next n−1
lines, each line contains three integers u,v,w(1≤u,v≤n,u≠v,1≤w≤109), denoting a bidirectional edge between vertex u and v with weight w
.
It is guaranteed that ∑n≤106
.
Output
For each test case, print a single line containing an integer, denoting the sum of all the edges' weight of MST(L(G))
.
Example
2
4
1 2 1
2 3 2
3 4 3
4
1 2 1
1 3 1
1 4 1
8
4
题解:题目给出一张图,让我们将每个边看成一个”点“,两“点”之间的权值为两边权之和。让我们找到这个“点”组成的图(题目命名为”线图“)的最小生成树的权值和。
我们可以从每条边权(即每个点的出边)的贡献入手,首先一个点的出边必须连通,否则构不成最小生成树。
那么对于特定的一个点,首先将其所有出边全部的权值加一遍,然后将其最小的一个边权乘以(这一点的度degree-2)即保证了最优解。(其实这样就是连degree-1条边使得保证最优解)
对于每一个点都这样,跑一遍即可。
#include<iostream>
#include<cstring>
#include<string>
#include<queue>
#include<stack>
#include<algorithm>
#include<stdio.h>
#include<map>
#include<set>
using namespace std;
typedef long long ll;
const int maxn=;
struct node
{
int v,w;
bool operator < (const node &r)const{
return w<r.w;
}
};
vector<node>G[maxn];
int main()
{
ios::sync_with_stdio();
int T;
cin>>T;
while(T--){
int n;
cin>>n;
for(int i=;i<=n;i++)G[i].clear();
for(int i=;i<=n-;i++){
int u,v,w;
cin>>u>>v>>w;
G[u].push_back((node){v,w});
G[v].push_back((node){u,w});
}
ll ans=;
for(int i=;i<=n;i++){
sort(G[i].begin(),G[i].end());
int minn=0x3f3f3f3f;
int degree=G[i].size();
for(int j=;j<degree;j++){
ans+=G[i][j].w;
minn=min(minn,G[i][j].w);
}
ans+=(ll)minn*(degree-);//当degree为1时与上面的ans相互消去
}
cout<<ans<<endl;
}
return ;
}
E - Minimum Spanning Tree Gym - 102220E (转化+贡献)的更多相关文章
- 【HDU 4408】Minimum Spanning Tree(最小生成树计数)
Problem Description XXX is very interested in algorithm. After learning the Prim algorithm and Krusk ...
- 数据结构与算法分析–Minimum Spanning Tree(最小生成树)
给定一个无向图,如果他的某个子图中,任意两个顶点都能互相连通并且是一棵树,那么这棵树就叫做生成树(spanning tree). 如果边上有权值,那么使得边权和最小的生成树叫做最小生成树(MST,Mi ...
- Educational Codeforces Round 3 E. Minimum spanning tree for each edge LCA/(树链剖分+数据结构) + MST
E. Minimum spanning tree for each edge Connected undirected weighted graph without self-loops and ...
- CF# Educational Codeforces Round 3 E. Minimum spanning tree for each edge
E. Minimum spanning tree for each edge time limit per test 2 seconds memory limit per test 256 megab ...
- Codeforces Educational Codeforces Round 3 E. Minimum spanning tree for each edge LCA链上最大值
E. Minimum spanning tree for each edge 题目连接: http://www.codeforces.com/contest/609/problem/E Descrip ...
- MST(Kruskal’s Minimum Spanning Tree Algorithm)
You may refer to the main idea of MST in graph theory. http://en.wikipedia.org/wiki/Minimum_spanning ...
- HDU 4408 Minimum Spanning Tree 最小生成树计数
Minimum Spanning Tree Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Ot ...
- [Educational Round 3][Codeforces 609E. Minimum spanning tree for each edge]
这题本来是想放在educational round 3的题解里的,但觉得很有意思就单独拿出来写了 题目链接:609E - Minimum spanning tree for each edge 题目大 ...
- Educational Codeforces Round 3 E. Minimum spanning tree for each edge 最小生成树+树链剖分+线段树
E. Minimum spanning tree for each edge time limit per test 2 seconds memory limit per test 256 megab ...
随机推荐
- pandas dataframe取差集:删掉已存在的数据,保留未插入的数据
适用场景: 插入数据到mysql中,中途中断,导致部分数据未插入成功.避免下次插入时插入了重复的数据. 思路: 1.读取已插入的数据, 2.读取全部数据(包含已插入和未插入的), 3.将已插入的数据添 ...
- 浅谈无字母数字构造webshell
0x00 问题 <?php include 'flag.php'; if(isset($_GET['code'])){ $code = $_GET['code']; if(strlen($cod ...
- POJ 1942:Paths on a Grid
Paths on a Grid Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 22918 Accepted: 5651 ...
- k8s安装helm
1.客户端安装 A.直接在github上下载二进制文件进行解压,下载地址:https://github.com/kubernetes/helm/releases B.将解压出来的二进制文件helm 拷 ...
- Cookie API和记录上次来访时间
1.什么是Cookie? Cookie是一种会话技术,用千将会话过程中的数据保存到用户的浏览器中,从而使浏览器和服务器可以更好地进行数据交互. 在现实生活中,当顾客在购物时,商城经常会赠送顾客一张会员 ...
- studentmanagement
package javatestywh; public class ScoreInformation { private String stunumber; private String name; ...
- 关于DSP仿真软件CCS中断点和探针的简单理解
关于DSP仿真软件CCS中断简单理解 (郑州大学姬祥老师编写) CCS中的2.0版本(实验所用)断点(Break Point) 和探针(Probe Point)之所以能组合使用,是因为我们在实现硬件仿 ...
- C/C++源程序到可执行程序的过程
源程序.cpp 预处理得到 预处理文件.i 编译得到 汇编文件.S 汇编得到 目标文件.o 链接得到 可执行文件 例子:main.cpp fun.cpp fun.h #inclu ...
- CodeForces - 350B(反向建图,)
B - Resort CodeForces - 350B B. Resort time limit per test 2 seconds memory limit per test 256 megab ...
- Java--类初始化
package httpclient.demo; public class StaticTest { public static void main(String[] args) { staticFu ...