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. 平方分割poj2104K-th Number

    K-th Number Time Limit: 20000MS   Memory Limit: 65536K Total Submissions: 59798   Accepted: 20879 Ca ...

  2. 状态压缩+枚举 UVA 11464 Even Parity

    题目传送门 /* 题意:求最少改变多少个0成1,使得每一个元素四周的和为偶数 状态压缩+枚举:枚举第一行的所有可能(1<<n),下一行完全能够由上一行递推出来,b数组保存该位置需要填什么 ...

  3. 牛客网-3 网易编程题(1拓扑&2二叉树的公共最近祖先&3快排找第K大数)

    1. 小明陪小红去看钻石,他们从一堆钻石中随机抽取两颗并比较她们的重量.这些钻石的重量各不相同.在他们们比较了一段时间后,它们看中了两颗钻石g1和g2.现在请你根据之前比较的信息判断这两颗钻石的哪颗更 ...

  4. js解析地址栏参数

    /** * 获取地址栏中url后面拼接的参数 * eg: * 浏览器地址栏中的地址:http://1.1.1.1/test.html?owner=2db08226-e2fa-426c-91a1-66e ...

  5. Python学习日记之中文支持

    解决中文输出错误 在开头添加 # -*- coding: utf-8 -*- 即可

  6. js中取整数的方法

    1.取整的方法 Math.floor( ) Math 对象的方法--取比当前数值小的最大整数(下取整). Math.ceil( ) Math对象的方法--取比当前数值大的最小整数(上取整). Math ...

  7. 【译】x86程序员手册22-6.4页级保护

    6.4 Page-Level Protection 页级保护 Two kinds of protection are related to pages: 与页相关的保护有两类: Restriction ...

  8. echarts之我用

    最近在用echarts做项目,抽点时间总结一下. 首先说一下什么是echarts.echarts是百度开发的类似于fusioncharts的图表展示控件.区别于fusioncharts的是echart ...

  9. Linux 中ifconfig和ip addr命令查看不到ip解决方法

    1.输入查看ip的命令ifconfig或ip addr,查不到ip 2.查看ens33网卡配置,输入 vi /etc/sysconfig/network-scripts/ifcfg-ens33 将ON ...

  10. QuickClip—界面原型设计

    1.需不需要设置用户登录/注册页? QuickClip没有提供该项功能.因为本产品为单纯的移动端视频编辑软件,是一个工具类软件.而且移动端软件本就追求的是方便快捷.简单易用,本产品不需要标识使用者的身 ...