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. Python天天学_04_基础四

    Python_day_04 金角大王: http://www.cnblogs.com/alex3714/articles/5765046.html ------Python是一个优雅的大姐姐 学习方式 ...

  2. 51 Nod 1069 Nim游戏

    分析: a1 xor a2 xor a3 ... xor an !=0 则为必胜态 a1 xor a2 xor a3 ... xor an ==0 则为必败态 也就是说只要计算异或值,如果非零则A赢, ...

  3. poj 2566 Bound Found 尺取法 变形

    Bound Found Time Limit: 5000MS   Memory Limit: 65536K Total Submissions: 2277   Accepted: 703   Spec ...

  4. LibreOJ NOI Round #2 Day 1

    LibreOJ NOI Round #2 Day 1 T1: 别被定义弄晕了 反着做,A->1/A+B 取倒数没法做,所以变成a/b,维护2*2的矩阵 区间?不用线段树,不用倍增 存在逆矩阵,直 ...

  5. 将Bean转换为Json形式的一个工具类

    这边遇到一个问题:  1.做一个bean类,实现一个函数,能够把bean生成json字符串.按字段作为key,字段值作为value的方式生成,并且按key的ascii码的升序生成.     2.提示: ...

  6. win10上的docker怎么设置开机不要自动启动 [问题点数:20分,结帖人xyq1986]

    次win开机都自动启动docker,感觉很耗资源,docker只是有时开发时需要用到,在docker的setting上的Start Docker Desktop when you log in取消了也 ...

  7. Python学习笔记:list的各种操作

    向一个列表中添加单个元素: my_list = []my_list.append('我爱你') 移除列表中的某个元素: my_list.pop(0) # 0是需要移除元素在列表中的index 或者是移 ...

  8. 1.zookeeper原理解析-数据存储之Zookeeper内存结构

    Zookeeper是怎么存储数据的,什么机制保证集群中数据是一致性,在网络异常,当机以及停电等异常情况下恢复数据的,我们知道数据库给我们提供了这些功能,其实zookeeper也实现了类似数据库的功能. ...

  9. redis深度历险:核心原理与应用实践--笔记

  10. 一、基础篇--1.1Java基础-Session和Cookie的区别【转】

    https://www.cnblogs.com/zlw-xf/p/8001383.html 1:cookie数据存放在客户的浏览器上(客户端),session数据放 @1:cookie不是很安全,别人 ...