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

Sample Output

0
17
16
26
#include<iostream>
#include<string>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
#define MAXN 55
#define INF 0x3f3f3f3f
bool been[MAXN];
int n,m,g[MAXN][MAXN],lowcost[MAXN];
int Prim(int beg)
{
memset(been,false,sizeof(been));
for(int i=;i<=n;i++)
{
lowcost[i] = g[beg][i];
}
been[beg] = true;
int ans = ;
for(int j=;j<n;j++)
{
int Minc = INF,k=-;
for(int i=;i<=n;i++)
{
if(!been[i]&&lowcost[i]<Minc)
{
Minc = lowcost[i];
k = i;
}
}
if(k==-) return -;
been[k] = true;
ans+=Minc;
for(int i=;i<=n;i++)
{
if(!been[i]&&lowcost[i]>g[k][i])
{
lowcost[i] = g[k][i];
}
}
}
return ans;
}
int main()
{
while(scanf("%d",&n),n)
{
scanf("%d",&m);
for(int i=;i<=n;i++)
for(int j=;j<=n;j++)
{
g[i][j] = INF;
}
for(int i=;i<m;i++)
{
int x,y,d;
scanf("%d%d%d",&x,&y,&d);
g[x][y] = min(g[x][y],d);
g[y][x] = min(g[y][x],d);
}
int ans = Prim();
cout<<ans<<endl;
}
}

最小生成树 B - Networking的更多相关文章

  1. (最小生成树) Networking -- POJ -- 1287

    链接: http://poj.org/problem?id=1287 Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 7494 ...

  2. 【转】并查集&MST题集

    转自:http://blog.csdn.net/shahdza/article/details/7779230 [HDU]1213 How Many Tables 基础并查集★1272 小希的迷宫 基 ...

  3. 【转载】图论 500题——主要为hdu/poj/zoj

    转自——http://blog.csdn.net/qwe20060514/article/details/8112550 =============================以下是最小生成树+并 ...

  4. Soj题目分类

    -----------------------------最优化问题------------------------------------- ----------------------常规动态规划 ...

  5. hdu图论题目分类

    =============================以下是最小生成树+并查集====================================== [HDU] 1213 How Many ...

  6. HDU图论题单

    =============================以下是最小生成树+并查集====================================== [HDU] 1213 How Many ...

  7. POJ 1287 Networking (最小生成树)

    Networking 题目链接: http://acm.hust.edu.cn/vjudge/contest/124434#problem/B Description You are assigned ...

  8. POJ 1287 Networking(最小生成树)

    题意  给你n个点 m条边  求最小生成树的权 这是最裸的最小生成树了 #include<cstdio> #include<cstring> #include<algor ...

  9. POJ 1287 Networking (最小生成树)

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

随机推荐

  1. 计算误差——ACM计算几何中的精度问题

    浮点数为何会有精度问题   占字节数 数值范围 十进制精度位数 float 4 -3.4e-38~3.4e38 6~7 double 8 -1.7e-308~1.7e308 14~15 如果内存不是很 ...

  2. Linux学习之路2 Bash的基本操作

    一.SHELL的介绍 shell分为两种:CLI(command Line Interface)和GUI(Graphical User Interface) 操作系统中的shell: GUI:GNOM ...

  3. ACM_求补集的交集

    求补集的交集 Time Limit: 2000/1000ms (Java/Others) Problem Description: 给定一个集合,然后再给出两个该集合的子集,求他们对应补集的交集. I ...

  4. ADB Usage Complete / ADB 用法大全

    ADB,即 Android Debug Bridge,它是 Android 开发/测试人员不可替代的强大工具,也是 Android 设备玩家的好玩具. 持续更新中,欢迎提 PR 和 Issue 补充指 ...

  5. MySQL与Sqlserver数据获取

    由于项目要求,一个.net mvc登录注册的东西网站必须放弃sqlserver数据去使用MySQL数据库,因此我遇到了一些问题,并找出相应的解决方法, 因为sqlserver跟MySQL的数据引擎不同 ...

  6. log4j建立propertie后要建立log4j2.xml

    log4j.properties ### \u8BBE\u7F6E### log4j.rootLogger = debug,stdout,D,E ### \u8F93\u51FA\u4FE1\u606 ...

  7. HTML增加附件

    <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/ ...

  8. Paxos,Raft,Zab一致性协议-Raft篇

    Raft是一个一致性算法,旨在易于理解.它提供了Paxos的容错和性能.不同之处在于它被分解为相对独立的子问题,它清楚地解决了实际系统所需的所有主要部分.我们希望Raft能够为更广泛的受众提供共识,并 ...

  9. 配置服务器 Ubuntu 记录+踩坑

    从零开始配置服务器用于ss+站点 1. SS 首先安装pyenv,安装pyenv之前先安装必要环境,具体命令行请见: https://github.com/pyenv/pyenv/wiki/Commo ...

  10. jmeter接口测试小结

    摘自:http://www.cnblogs.com/houzhizhe/p/6839736.html JMeter做http接口压力测试 测前准备 用JMeter做接口的压测非常方便,在压测之前我们需 ...