Description

Bessie has been hired to build a cheap internet network among Farmer John's N (2 <= N <= 1,000) barns that are conveniently numbered 1..N. FJ has already done some surveying, and found M (1 <= M <= 20,000) possible connection routes between pairs of barns. Each possible connection route has an associated cost C (1 <= C <= 100,000). Farmer John wants to spend the least amount on connecting the network; he doesn't even want to pay Bessie.

Realizing Farmer John will not pay her, Bessie decides to do the worst job possible. She must decide on a set of connections to install so that (i) the total cost of these connections is as large as possible, (ii) all the barns are connected together (so that it is possible to reach any barn from any other barn via a path of installed connections), and (iii) so that there are no cycles among the connections (which Farmer John would easily be able to detect). Conditions (ii) and (iii) ensure that the final set of connections will look like a "tree".

Input

* Line 1: Two space-separated integers: N and M

* Lines 2..M+1: Each line contains three space-separated integers A, B, and C that describe a connection route between barns A and B of cost C.

Output

* Line 1: A single integer, containing the price of the most expensive tree connecting all the barns. If it is not possible to connect all the barns, output -1.

Sample Input

5 8
1 2 3
1 3 7
2 3 10
2 4 4
2 5 8
3 4 6
3 5 2
4 5 17

Sample Output

42

Hint

OUTPUT DETAILS:

The most expensive tree has cost 17 + 8 + 10 + 7 = 42. It uses the following connections: 4 to 5, 2 to 5, 2 to 3, and 1 to 3.

 
 
题目意思:为了防止雇主不给钱或者少给钱,安装网络的人想要是每一个谷堆之间连通网络的成本最大,即生成一个最大生成树。
 
 
 
\\\克鲁斯卡尔算法
 
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int n,m,sum;
struct node
{
int start;///起点
int end;///终点
int power;///权值
} edge[20050];
int pre[20050];
int cmp(node a,node b)
{
return a.power<b.power;///按权值降序排列
// return a.power<b.power
}
int find(int x)///并查集找祖先
{
int a;///循环法
a=x;
while(pre[a]!=a)
{
a=pre[a];
}
return a;
}
void merge(int x,int y,int n)
{
int fx =find(x);
int fy =find(y);
if(fx!=fy)
{
pre[fx]=fy;
sum+=edge[n].power;
}
}
int main()
{
int i,x,count;
while(scanf("%d%d",&n,&m)!=EOF)
{
sum=0;
count=0;
for(i=1; i<=m; i++)
{
scanf("%d%d%d",&edge[i].start,&edge[i].end,&x);
edge[i].power=x;
//edge[i].power=-x;
}
for(i=1; i<=m; i++) ///并查集的初始化
{
pre[i]=i;
}
sort(edge+1,edge+m+1,cmp);
for(i=1; i<=m; i++)
{
merge(edge[i].start,edge[i].end,i);
}
for(i=1; i<=n; i++)
{
if(pre[i]==i)
{
count++;
}
}
if(count==1)
{
printf("%d\n",sum);
// printf("%d\n",-sum);
}
else
{
printf("-1\n");
}
}
return 0;
}

  

///普里姆算法

#include<stdio.h>
#include<string.h>
#define MAX 0x3f3f3f3f
using namespace std;
int logo[1010];///用0和1来表示是否被选择过
int map1[1010][1010];
int dis[1010];///记录任意一点到这一点的最近的距离
int n,m;
int prim()
{
int i,j,now;
int sum=0;
for(i=1; i<=n; i++) ///初始化
{
dis[i]=MAX;
logo[i]=0;
}
for(i=1; i<=n; i++)
{
dis[i]=map1[1][i];
}
dis[1]=0;
logo[1]=1;
for(i=1; i<n; i++) ///循环查找
{
now=-MAX;
int max1=-MAX;
for(j=1; j<=n; j++)
{
if(logo[j]==0&&dis[j]>max1)
{
now=j;
max1=dis[j];
}
}
if(now==-MAX)///防止不成图
{
break;
}
logo[now]=1;
sum=sum+max1;
for(j=1; j<=n; j++) ///填入新点后更新最小距离,到顶点集的距离
{
if(logo[j]==0&&dis[j]<map1[now][j])
{
dis[j]=map1[now][j];
}
}
}
if(i<n)
{
printf("-1\n");
}
else
{
printf("%d\n",sum);
}
}
int main()
{
int i,j;
int a,b,c;
while(scanf("%d%d",&n,&m)!=EOF)///n是点数
{
for(i=1; i<=n; i++)
{
for(j=i; j<=n; j++)
{
if(i==j)
{
map1[i][j]=map1[j][i]=0;
}
else
{
map1[i][j]=map1[j][i]=-MAX;
}
}
}
for(i=0; i<m; i++)
{
scanf("%d%d%d",&a,&b,&c);
if(map1[a][b]<c)///防止出现重边
{
map1[a][b]=map1[b][a]=c;
}
}
prim();
}
return 0;
}

  

Bad Cowtractors(最大生成树)的更多相关文章

  1. [POJ2377]Bad Cowtractors(最大生成树,Kruskal)

    题目链接:http://poj.org/problem?id=2377 于是就找了一道最大生成树的AC了一下,注意不连通的情况啊,WA了一次. /* ━━━━━┒ギリギリ♂ eye! ┓┏┓┏┓┃キリ ...

  2. 杭电ACM分类

    杭电ACM分类: 1001 整数求和 水题1002 C语言实验题——两个数比较 水题1003 1.2.3.4.5... 简单题1004 渊子赛马 排序+贪心的方法归并1005 Hero In Maze ...

  3. 转载:hdu 题目分类 (侵删)

    转载:from http://blog.csdn.net/qq_28236309/article/details/47818349 基础题:1000.1001.1004.1005.1008.1012. ...

  4. poj图论解题报告索引

    最短路径: poj1125 - Stockbroker Grapevine(多源最短路径,floyd) poj1502 - MPI Maelstrom(单源最短路径,dijkstra,bellman- ...

  5. POJ - 2377 Bad Cowtractors Kru最大生成树

    Bad Cowtractors Bessie has been hired to build a cheap internet network among Farmer John's N (2 < ...

  6. BZOJ 3390: [Usaco2004 Dec]Bad Cowtractors牛的报复(最大生成树)

    这很明显就是最大生成树= = CODE: #include<cstdio>#include<iostream>#include<algorithm>#include ...

  7. poj 2377 Bad Cowtractors (最大生成树prim)

    Bad Cowtractors Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 131072/65536K (Java/Other) To ...

  8. bzoj 3390: [Usaco2004 Dec]Bad Cowtractors牛的报复 -- 最大生成树

    3390: [Usaco2004 Dec]Bad Cowtractors牛的报复 Time Limit: 1 Sec  Memory Limit: 128 MB Description     奶牛贝 ...

  9. bzoj 3390: [Usaco2004 Dec]Bad Cowtractors牛的报复【最大生成树】

    裸的最大生成树,注意判不连通情况 #include<iostream> #include<cstdio> #include<algorithm> using nam ...

随机推荐

  1. 正则表达式-Regular expression学习笔记

    正则表达式 正则表达式(Regular expression)是一种符号表示法,被用来识别文本模式. 最近在学习正则表达式,今天整理一下其中的一些知识点 grep - 打印匹配行 grep 是个很强大 ...

  2. js扩展String.prototype.format字符串拼接的功能

    1.题外话,有关概念理解:String.prototype 属性表示 String原型对象.所有 String 的实例都继承自 String.prototype. 任何String.prototype ...

  3. JQuery制作网页——第九章 表单验证

    1.  表单验证:减轻服务器的压力.保证输入的数据符合要求: 2.  常用的表单验证:日期格式.表单元素是否为空.用户名和密码.E-mail地址.身份证号码等: 3.  表单验证的思路: 1.     ...

  4. 开发和调试第一个 LLVM Pass

    1. 下载和编译 LLVM LLVM 下载地址 http://releases.llvm.org/download.html,目前最新版是 6.0.0,下载完成之后,执行 tar 解压 llvm 包: ...

  5. angularjs中 $watch 和$on 2种监听的区别?

    1.$watch简单使用 $watch是一个scope函数,用于监听模型变化,当你的模型部分发生变化时它会通知你. $watch(watchExpression, listener, objectEq ...

  6. 一分钟完成pip安装

    很多实用Python的小伙伴都需要使用pip安装相应的包,对于初学者而已,检查遇到pip安装不成功的情况,如以下典型错误: Traceback (most recent call last): Fil ...

  7. Hadoop入门学习路线

    走上大数据的自学之路....,Hadoop是走上大数据开发学习之路的第一个门槛. Hadoop,是Apache的一个开源项目,开发人员可以在不了解分布式底层细节,开发分布式程序,充分利用集群进行高速运 ...

  8. arping命令用法

    arping命令使用说明 BusyBox v1.17.3 (2011-07-20 17:01:30 CST) multi-call binary. Usage: arping [-fqbDUA] [- ...

  9. UDP server Code

    Code Example: The following programs demonstrate the use of getaddrinfo(), gai_strerror(), freeaddri ...

  10. 成都Uber优步司机奖励政策(1月27日)

    滴快车单单2.5倍,注册地址:http://www.udache.com/ 如何注册Uber司机(全国版最新最详细注册流程)/月入2万/不用抢单:http://www.cnblogs.com/mfry ...