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. 我在做评论功能时学到的js一些思路

    在提交评论的时候,如何判断是一级评论还是二级评论(因为都是通过一个文本域提交评论),思路:声明一个全局变量,如果是回复(二级评论)那么会触发点击回复事件,在这个事件的回调函数里给全局变量设置为true ...

  2. Arthas--Java在线分析诊断工具(阿尔萨斯)

    序言 Arthas是一款阿里巴巴开源的 Java 线上诊断工具,功能非常强大,可以解决很多线上不方便解决的问题. 资料 https://blog.csdn.net/youanyyou/article/ ...

  3. maven项目创建5 service层整合

    创建service相关文件 创建applicationContext-service.xml文件 <?xml version="1.0" encoding="UTF ...

  4. Nowcoder 练习赛 17 C 操作数 ( k次前缀和、矩阵快速幂打表找规律、组合数 )

    题目链接 题意 :  给定长度为n的数组a,定义一次操作为: 1. 算出长度为n的数组s,使得si= (a[1] + a[2] + ... + a[i]) mod 1,000,000,007: 2. ...

  5. 翻译一篇英文文章,主要是给自己看的——在ASP.NET Core Web Api中如何刷新token

    原文地址 :https://www.blinkingcaret.com/2018/05/30/refresh-tokens-in-asp-net-core-web-api/ 先申明,本人英语太菜,每次 ...

  6. 【Spark机器学习速成宝典】模型篇06随机森林【Random Forests】(Python版)

    目录 随机森林原理 随机森林代码(Spark Python) 随机森林原理 参考:http://www.cnblogs.com/itmorn/p/8269334.html 返回目录 随机森林代码(Sp ...

  7. apache源码安装 转载

    转载 1.先进入/usr/local/中创建三个文件夹 apr apr-util apache cd /usr/local目录 mkdir apr mkdir apr-util mkdir apach ...

  8. [转]maven中scope详解

    在POM 4中,<dependency>中还引入了<scope>,它主要管理依赖的部署.目前<scope>可以使用5个值: * compile,缺省值,适用于所有阶 ...

  9. k8s, etcd 多节点集群部署问题排查记录

    目录 文章目录 目录 部署环境 1. etcd 集群启动失败 解决 2. etcd 健康状态检查失败 解决 3. kube-apiserver 启动失败 解决 4. kubelet 启动失败 解决 5 ...

  10. 【DVWA】File Upload(文件上传漏洞)通关教程

    日期:2019-08-01 17:28:33 更新: 作者:Bay0net 介绍: 0x01. 漏洞介绍 在渗透测试过程中,能够快速获取服务器权限的一个办法. 如果开发者对上传的内容过滤的不严,那么就 ...