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<algorithm>
using namespace std;
const int N = 55, M = 3000;
int par[N], n, m, ans;
struct edge{int u, v, w;} e[M];
bool cmp(edge a, edge b){return a.w < b.w;} int Find(int x)
{
int r = x, tmp;
while(par[r] >= 0) r = par[r];
while(x != r)
{
tmp = par[x];
par[x] = r;
x = tmp;
}
return r;
} void Union (int u, int v)
{
int ru = Find(u), rv = Find(v), tmp = par[ru] + par[rv];
if(par[ru] < par[rv])
par[rv] = ru, par[ru] = tmp;
else
par[ru] = rv, par[rv] = tmp;
} void kruskal()
{
memset(par, -1, sizeof(par));
int cnt = 0;
for(int i = 1; i <= m; ++i)
{
int u = e[i].u, v = e[i].v;
if(Find(u) != Find(v))
{
++cnt;
ans += e[i].w;
Union(u, v);
}
if(cnt >= n - 1) break;
}
} int main()
{
int u, v, w;
while(scanf("%d", &n), n)
{
scanf("%d", &m);
for(int i = 1; i <= m; ++i)
{
scanf("%d%d%d", &u, &v, &w);
e[i].u = u, e[i].v = v, e[i].w = w;
}
sort(e + 1, e + m + 1, cmp);
ans = 0;
kruskal();
printf("%d\n", ans);
}
return 0;
}

Networking


Time Limit: 2 Seconds      Memory Limit: 65536 KB


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

POJ 1287 Networking(最小生成树)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. 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 ...

  5. 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 ...

  6. poj 1287 Networking【最小生成树prime】

    Networking Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 7321   Accepted: 3977 Descri ...

  7. [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 ...

  8. POJ - 1287 Networking 【最小生成树Kruskal】

    Networking Description You are assigned to design network connections between certain points in a wi ...

  9. POJ 1287 Networking(最小生成树裸题有重边)

    Description You are assigned to design network connections between certain points in a wide area. Yo ...

随机推荐

  1. 默认情况下,不使用of子句表示在select所有的数据表中加锁(转)

    Select …forupdate语句是我们经常使用手工加锁语句.通常情况下,select语句是不会对数据加锁,妨碍影响其他的DML和DDL操作.同时,在多版本一致读机制的支持下,select语句也不 ...

  2. Processing.js

    Processing.js Processing.js 1.4.1 released!

  3. 【读书笔记】《未来闪影》罗伯特·J·索耶

    真是一本引人入胜的书! 看了不到一半,就有一种置身其中的感觉,要是我也能看到自己二十年后的生活,哪怕只有1分43秒,该是一件多么奇妙的事情.但忧虑也随之而来,如果二十年后我没有成为现在想成为的人,现在 ...

  4. C++智能指针--weak_ptr

    weak_ptr是对对象的一种弱引用,它不会添加对象的引用计数.weak_ptr和shared_ptr之间能够相互转换.shared_ptr能够直接赋值给week_ptr,week_ptr可通过调用l ...

  5. BAD packet signature 18245 错误解决

    1.错误信息 2014-7-15 2:46:38 org.apache.jk.common.MsgAjp processHeader 严重: BAD packet signature 18245 20 ...

  6. poj 1743 男人八题之后缀数组求最长不可重叠最长重复子串

    Musical Theme Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 14874   Accepted: 5118 De ...

  7. JNI生成so

    软件:android-ndk-r8 推荐使用这个版本,可以直接不用安装Cygwin软件就可以编译. 然后在系统环境变量中path选项中添加安装路径,比如我的:C:\android-ndk-r8: 然后 ...

  8. 主流JavaScript框架(Dojo、Google Closure、jQuery、Prototype、Mootools和YUI)的分析和对比

    本文主要选取了目前比较流行的JavaScript框架Dojo.Google Closure.jQuery.Prototype.Mootools和YUI进行对比,主要是根据网上的资料整理而成,希望可以供 ...

  9. 关于NSArray的几种排序:

    #利用数组的sortedArrayUsingComparator调用 NSComparator  当中NSComparator事实上就是一个返回NSComparisonResult的block. ty ...

  10. Linux 技巧之 Grub 超实用技巧

    1. 简单介绍 什么是 GRUB?GRUB 全名Grand Unified Boot Loader,它是一个引导装入器 -- 它负责装入内核并引导 Linux 系统.GRUB 还能够引导其他操作系统, ...