You are assigned to design network connections between certain points in a wide area. You are given a set of points in the area, and a set of possible routes for the cables that may connect pairs of points. For each possible route between two points, you are given the length of the cable that is needed to connect the points over that route. Note that there may exist many possible routes between two given points. It is assumed that the given possible routes connect (directly or indirectly) each two points in the area.
Your task is to design the network for the area, so that there is a connection (direct or indirect) between every two points (i.e., all the points are interconnected, but not necessarily by a direct cable), and that the total length of the used cable is minimal.

Input

The input file consists of a number of data sets. Each data set defines one required network. The first line of the set contains two integers: the first defines the number P of the given points, and the second the number R of given routes between the points. The following R lines define the given routes between the points, each giving three integer numbers: the first two numbers identify the points, and the third gives the length of the route. The numbers are separated with white spaces. A data set giving only one number P=0 denotes the end of the input. The data sets are separated with an empty line.
The maximal number of points is 50. The maximal length of a given route is 100. The number of possible routes is unlimited. The nodes are identified with integers between 1 and P (inclusive). The routes between two points i and j may be given as i j or as j i.

Output

For each data set, print one number on a separate line that gives the total length of the cable used for the entire designed network.

Sample Input

1 0

2 3
1 2 37
2 1 17
1 2 68 3 7
1 2 19
2 3 11
3 1 7
1 3 5
2 3 89
3 1 91
1 2 32 5 7
1 2 5
2 3 7
2 4 8
4 5 11
3 5 10
1 5 6
4 2 12 0
#include<iostream>
#include<algorithm>
using namespace std;
int ans=,tot=;
const int N = 1e5;
int f[];
struct ac{
int v,u,w;
}edge[N];
bool cmp(ac a,ac b){
return a.w<b.w;
}
inline int find(int x){
if(x!=f[x])f[x]=find(f[x]);return f[x];
}
inline int join(int x,int y,int w){
int fx=find(x),fy=find(y);
if(fx!=fy){
f[fx]=fy;ans+=w;tot++;
}
}
int main()
{
int n;string a,b;int m,w,cnt=;
while(cin>>n>>m){
ans=tot=cnt=;
for(int i = ;i <=n;++i)f[i]=i;
for(int i = ;i < m;++i){
cin>>edge[i].u>>edge[i].v>>edge[i].w;
}
sort(edge,edge+m,cmp);
for(int i = ;i < m;++i){
join(edge[i].u,edge[i].v,edge[i].w);
if(tot==n-)break;
}
cout<<ans<<endl;
}
return ;
}

POJ - 1287 Networking (最小生成树&并查集的更多相关文章

  1. POJ 1287 Networking (最小生成树)

    Networking Time Limit:1000MS     Memory Limit:10000KB     64bit IO Format:%I64d & %I64u Submit S ...

  2. POJ 1287 Networking (最小生成树模板题)

    Description You are assigned to design network connections between certain points in a wide area. Yo ...

  3. ZOJ1372 POJ 1287 Networking 网络设计 Kruskal算法

    题目链接:problemCode=1372">ZOJ1372 POJ 1287 Networking 网络设计 Networking Time Limit: 2 Seconds     ...

  4. UVA 1395 苗条的生成树(最小生成树+并查集)

    苗条的生成树 紫书P358 这题最后坑了我20分钟,怎么想都对了啊,为什么就wa了呢,最后才发现,是并查集的编号搞错了. 题目编号从1开始,我并查集编号从0开始 = = 图论这种题真的要记住啊!!题目 ...

  5. POJ.1287 Networking (Prim)

    POJ.1287 Networking (Prim) 题意分析 可能有重边,注意选择最小的边. 编号依旧从1开始. 直接跑prim即可. 代码总览 #include <cstdio> #i ...

  6. CSP 201703-4 地铁修建【最小生成树+并查集】

    问题描述 试题编号: 201703-4 试题名称: 地铁修建 时间限制: 1.0s 内存限制: 256.0MB 问题描述: 问题描述 A市有n个交通枢纽,其中1号和n号非常重要,为了加强运输能力,A市 ...

  7. 关于最小生成树(并查集)prime和kruskal

    适合对并查集有一定理解的人.  新手可能看不懂吧.... 并查集简单点说就是将相关的2个数字联系起来 比如 房子                      1   2    3   4  5   6 ...

  8. poj 1733(带权并查集+离散化)

    题目链接:http://poj.org/problem?id=1733 思路:这题一看就想到要用并查集做了,不过一看数据这么大,感觉有点棘手,其实,我们仔细一想可以发现,我们需要记录的是出现过的节点到 ...

  9. poj 1182 食物链 (并查集)

    http://poj.org/problem?id=1182 关于并查集 很好的一道题,开始也看了一直没懂.这次是因为<挑战程序设计竞赛>书上有讲解看了几遍终于懂了.是一种很好的思路,跟网 ...

  10. POJ 1182 食物链(并查集拆点)

    [题目链接] http://poj.org/problem?id=1182 [题目大意] 草原上有三种物种,分别为A,B,C A吃B,B吃C,C吃A. 1 x y表示x和y是同类,2 x y表示x吃y ...

随机推荐

  1. eclipse切换工作空间

  2. k8s删除节点后再重新添加进去(踩坑)

    开启本地集群,发现一台节点出问题了,想删除再换一台节点,结果就踩坑了,还好本地有好几套环境. 再master节点执行以下命令 [root@k8s-master ~]# kubectl drain k8 ...

  3. python3.6+Xadmin2.0系列(一) xadmin下载及安装

    环境配置:win7+python3.6+Django2.1+xadmin2+PyCharm 一.Xadmin下载及安装: 1.下载: 下载地址:https://github.com/sshwsfc/x ...

  4. less基本用法:持续归纳中

    todo 1,嵌套语法:https://www.w3cschool.cn/less/nested_directives_bubbling.html 简单来说就是可以与html一样去写css,并且会继承 ...

  5. windos下hosts文件

    win+R c:\windows\system32\drivers\etc

  6. debug1: Could not open authorized keys

    ssh登录的时候一直日志一直出现debug1: Could not open authorized keys登录不上,检查文件夹权限都正常用这条命令解决了 restorecon -FRvv /home ...

  7. CondenseNet: An Efficient DenseNet using Learned Group Convolutions

    1. 摘要 作者提出了一个前所未有高效的新奇网络结构,称之为 CondenseNet,该结构结合了密集连接性和可学习的分组卷积模块. 密集连接性有利于网络中的特征复用,而可学习的分组卷积模块则可以移除 ...

  8. IDEA打开最近打开的项目以及关闭项目

    关闭的是当前项目

  9. 如何解决错误【selenium.common.exceptions.SessionNotCreatedException】

    如何解决错误[selenium.common.exceptions.SessionNotCreatedException]   [问题起因] 2018年12月26日晚,启动我的pycharm准备学习s ...

  10. Reactjs之静态路由、动态路由以及Get传值以及获取

    1.新增知识点 /* react路由的配置: 1.找到官方文档 https://reacttraining.com/react-router/web/example/basic 2.安装 cnpm i ...