Networking
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 6432   Accepted: 3488

Description

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

Source

这道题其实就是求mst,贴AC代码:


#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std; const int MAXN=55;
const int MAXM=1e6;
int father[MAXN];
struct Edge
{
int u,v,w;
}edge[MAXM];
int tot;
void addedge(int u,int v,int w)
{
edge[tot].u=u;
edge[tot].v=v;
edge[tot++].w=w;
}
int find_set(int x)
{
if(father[x]==-1)
return x;
else
return father[x]=find_set(father[x]);
}
bool cmp(Edge x,Edge y)
{
return x.w<y.w;
}
int Kruskal(int n)
{
memset(father,-1,sizeof(father));
sort(edge,edge+tot,cmp);
int ans=0;
int num=0;
for(int i=0;i<tot;i++)
{
int u=edge[i].u;
int v=edge[i].v;
int w=edge[i].w;
int fa1=find_set(u);
int fa2=find_set(v);
if(fa1!=fa2)
{
ans+=w;
num++;
father[fa1]=fa2;
}
if(num==n-1)
break;
}
if(num<n-1)
return -1;
else
return ans;
}
int main()
{
int n,m;
while(scanf("%d%d",&n,&m))
{
if(n==0)
break;
tot=0;
int u,v,w;
for(int i=0;i<m;i++)
{
scanf("%d%d%d",&u,&v,&w);
u--;
v--;
addedge(u,v,w);
}
int ans=Kruskal(n);
printf("%d\n",ans);
}
return 0;
}

POJ_1287_mst的更多相关文章

随机推荐

  1. yum命令

    1)查询 yum   list #查询所有可用软件包列表 yum search 关键字  #搜索服务器上和关键字相关的包 #对比rpm的查询:rpm  -qa | grep xxx 2)安装 yum ...

  2. 【BZOJ2013】【JSOI2008】球形空间产生器

    看chty代码 原题: BZOJ挂了--等好了补上题面 有一个球形空间产生器能够在n维空间中产生一个坚硬的球体.现在,你被困在了这个n维球体中,你只知道球面上n+1个点的坐标,你需要以最快的速度确定这 ...

  3. 黑马程序员——JAVA基础之主函数main和静态static,静态代码块

    ------- android培训.java培训.期待与您交流! ---------- 主函数:是一个特殊的函数.作为程序的入口,可以被jvm调用. 主函数的定义: public:代表着该函数访问权限 ...

  4. GOOGLE的专业使用方法(转)

    搜索引擎命令大全! 1.双引号 把搜索词放在双引号中,代表完全匹配搜索,也就是说搜索结果返回的页面包含双引号中出现的所有的词,连顺序也必须完全匹配.bd和Google 都支持这个指令.例如搜索: “s ...

  5. 【转】NSString属性什么时候用copy,什么时候用strong?

    原文网址:http://www.cocoachina.com/ios/20150512/11805.html 我们在声明一个NSString属性时,对于其内存相关特性,通常有两种选择(基于ARC环境) ...

  6. 整合UMDH结果的一个小工具

    ua.exe使用方法: 1.将UMDH生成的logcompare.txt改名为1.txt,内容示例: // Debug library initialized ... DBGHELP: moxia_d ...

  7. 【C++11】30分钟了解C++11新特性

    作者:王选易,出处:http://www.cnblogs.com/neverdie/ 欢迎转载,也请保留这段声明.如果你喜欢这篇文章,请点[推荐].谢谢! 什么是C++11 C++11是曾经被叫做C+ ...

  8. 大白话系列之C#委托与事件讲解(三)

    今天我接着上面的3篇文章来讲一下,为什么我们在日常的编程活动中遇到这么多sender,EventArgs e 参数:protected void Page_Load(object sender, Ev ...

  9. ABBYY PDF Transformer+ Pro支持全世界189种语言

    ABBYY PDF Transformer+ Pro版支持189种语言,包括我们人类的自然语言.人造语言以及正式语言.受支持的语言可能会因产品的版本不同而各异.本文具体列举了所有ABBYY PDF T ...

  10. unity中的欧拉角

    unity中欧拉角用的是heading - pitch -bank系统(zxy惯性空间旋转系统):当认为旋转顺序是zxy时,是相对于惯性坐标系旋转.当认为旋转顺序是yxz时,是相对于物体坐标系旋转. ...