POJ 1287 Networking(最小生成树裸题有重边)
Description
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 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
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
Sample Output
0
17
16
26
#include<iostream>
#include<cstdio>
#include<queue>
#include<vector>
#include<cstring>
#include<string>
#include<algorithm>
#include<map>
#include<cmath>
#include<math.h>
using namespace std; int fa[]; struct node
{
int u,v,c;
}a[]; bool cmp(node b,node d)
{
return b.c<d.c;
} int find(int x)
{
return fa[x]==x ? x:fa[x]=find(fa[x]);
} int main()
{ int n,m,ans;
while(~scanf("%d",&n)&&n)
{
ans=;
scanf("%d",&m);
for(int i=;i<m;i++)
{
scanf("%d%d%d",&a[i].u,&a[i].v,&a[i].c);
}
sort(a,a+m,cmp);
for(int i=;i<=n;i++)
fa[i]=i;
for(int i=;i<m;i++)
{
int p=find(a[i].u),q=find(a[i].v);
//因为排过序了所以有重边也是直接取小的
if(p!=q)
{
fa[p]=q;
ans+=a[i].c;
}
}
printf("%d\n",ans); }
return ;
}
//直接拿前面一篇博客改的代码
POJ 1287 Networking(最小生成树裸题有重边)的更多相关文章
- POJ 1258 + POJ 1287 【最小生成树裸题/矩阵建图】
Farmer John has been elected mayor of his town! One of his campaign promises was to bring internet c ...
- 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 ...
- POJ 1287 Networking (最小生成树)
Networking Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64u Submit S ...
- BZOJ 1083 [SCOI2005]繁忙的都市 (最小生成树裸题无重边) 超简单写法!!
Description 城市C是一个非常繁忙的大都市,城市中的道路十分的拥挤,于是市长决定对其中的道路进行改造.城市C的道路是这样分布的:城市中有n个交叉路口,有些交叉路口之间有道路相连,两个交叉路口 ...
- 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 ...
- 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 ...
- [kuangbin带你飞]专题六 最小生成树 POJ 1287 Networking
最小生成树模板题 跑一次kruskal就可以了 /* *********************************************** Author :Sun Yuefeng Creat ...
- HDU 1102 最小生成树裸题,kruskal,prim
1.HDU 1102 Constructing Roads 最小生成树 2.总结: 题意:修路,裸题 (1)kruskal //kruskal #include<iostream> ...
- POJ 3468 线段树裸题
这些天一直在看线段树,因为临近期末,所以看得断断续续,弄得有些知识点没能理解得很透切,但我也知道不能钻牛角尖,所以配合着刷题来加深理解. 然后,这是线段树裸题,而且是最简单的区间增加与查询,我参考了A ...
随机推荐
- Markdown 绘制 UML 图 -- PlantUML + Gravizo(转)
原文地址:Markdown 绘制 UML 图 -- PlantUML + Gravizo
- Python---字典常用方法总结
字典是一种key-value的数据类型,字典里必须写Key和value,字典的优点是取数方便和速度快.字典的特性: 1.字典是无序的,因为它没有下标,用key来当索引,所以是无序的 2.字典的key必 ...
- hdu多校1004 Distinct Values
Distinct Values Time Limit: / MS (Java/Others) Memory Limit: / K (Java/Others) Total Submission(s): ...
- PHP如何自定义PHP内置函数
其实对于PHP程序员,有个纯PHP的解决方案.在php.ini里有个配置项 auto_prepend_file,可以设置一个PHP文件作为每次执行前自动加载的文件. 在这个文件里写函数,你就可以当成定 ...
- Qt中漂亮的几款QSS
/* === Shared === */QStackedWidget, QLabel, QPushButton, QRadioButton, QCheckBox, QGroupBox, QStatus ...
- Android--Android Studio 打开ADM报错
Android studio无法打开类似与eclipse的DDMS, 在android studio里点击android device monitor(点击菜单栏里的Tools->Android ...
- C++11 并发之std::thread std::mutex
https://www.cnblogs.com/whlook/p/6573659.html (https://www.cnblogs.com/lidabo/p/7852033.html) C++:线程 ...
- 【Oracle安装卸载】oracle卸载
Oracle卸载比较麻烦,不能简单卸载就完成了,有时没有卸载完整,下次安装不能很好的安装: 当然Oracle卸载也没有那么难,只是步骤比较多.Oracle10g还是Oracle11g卸载步骤都是一样的 ...
- java 一些容易忽视的小点-数据类型和运算符篇
注释 文档注释: 以"/**"开头以"*/"结尾,注释中包含一些说明性的文字及一些JavaDoc标签(后期写项目时,可以生成项目的API) 行注释: 以 ...
- Jdbc连接Oracle12C集群环境
jdbc.url=jdbc:Oracle:thin:@(DESCRIPTION =(ADDRESS_LIST =(ADDRESS = (PROTOCOL = TCP)(HOST = 192.31.0. ...