POJ 1861 & ZOJ 1542 Network(最小生成树之Krusal)
2024-08-25 01:29:03
题目链接:
PKU:http://poj.org/problem?id=1861
ZJU:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=542
Description
Andrew is working as system administrator and is planning to establish a new network in his company. There will be N hubs in the company, they can be connected to each other using cables. Since each worker of the company must have access to the whole network,
each hub must be accessible by cables from any other hub (with possibly some intermediate hubs).
Since cables of different types are available and shorter ones are cheaper, it is necessary to make such a plan of hub connection, that the maximum length of a single cable is minimal. There is another problem — not each hub can be connected to any other one
because of compatibility problems and building geometry limitations. Of course, Andrew will provide you all necessary information about possible hub connections.
You are to help Andrew to find the way to connect hubs so that all above conditions are satisfied.
each hub must be accessible by cables from any other hub (with possibly some intermediate hubs).
Since cables of different types are available and shorter ones are cheaper, it is necessary to make such a plan of hub connection, that the maximum length of a single cable is minimal. There is another problem — not each hub can be connected to any other one
because of compatibility problems and building geometry limitations. Of course, Andrew will provide you all necessary information about possible hub connections.
You are to help Andrew to find the way to connect hubs so that all above conditions are satisfied.
Input
The first line of the input contains two integer numbers: N - the number of hubs in the network (2 <= N <= 1000) and M - the number of possible hub connections (1 <= M <= 15000). All hubs are numbered from 1 to N. The following M lines contain information about
possible connections - the numbers of two hubs, which can be connected and the cable length required to connect them. Length is a positive integer number that does not exceed 106. There will be no more than one way to connect two hubs. A hub cannot
be connected to itself. There will always be at least one way to connect all hubs.
possible connections - the numbers of two hubs, which can be connected and the cable length required to connect them. Length is a positive integer number that does not exceed 106. There will be no more than one way to connect two hubs. A hub cannot
be connected to itself. There will always be at least one way to connect all hubs.
Output
Output first the maximum length of a single cable in your hub connection plan (the value you should minimize). Then output your plan: first output P - the number of cables used, then output P pairs of integer numbers - numbers of hubs connected by the corresponding
cable. Separate numbers by spaces and/or line breaks.
cable. Separate numbers by spaces and/or line breaks.
Sample Input
4 6
1 2 1
1 3 1
1 4 2
2 3 1
3 4 1
2 4 1
Sample Output
1
4
1 2
1 3
2 3
3 4
Source
Northeastern Europe 2001, Northern Subregion
题意:
有n个顶点,m条边,每条边都是双向的,而且有一定的长度。要求使每一个顶点都连通,而且要使总长度最短,
输出最大边、边的总数和所选择的边。
PS:
貌似题目的案例有点问题,卡了好久!
应该输出的是:
1
3
1 3
2 3
2 4
代码例如以下:
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn = 15017;
int father[maxn];
struct edge
{
int x,y,v;
};
struct edge ed[maxn],ansa[maxn]; bool cmp(edge a,edge b)
{
return a.v<b.v;
} int find(int x)
{
if(x==father[x])
return x;
return father[x]=find(father[x]);
} void Krusal(int n,int m)
{
int i,fx,fy,cnt;
int ans=0;
for(i = 1; i <= n; i++)
father[i]=i;
sort(ed,ed+m,cmp);//对边的排序
cnt=0;
int max=-1;
for(i=0; i<m; i++)
{
fx=find(ed[i].x);
fy=find(ed[i].y);
if(fx!=fy)
{
ans+=ed[i].v;
father[fx]=fy;
ansa[cnt].x=ed[i].x;
ansa[cnt++].y=ed[i].y;
if(max<ed[i].v)
max=ed[i].v;
}
}
printf("%d\n%d\n",max,cnt);
for(i=0; i<cnt; i++)
printf("%d %d\n",ansa[i].x,ansa[i].y);
} int main()
{
int t;
int n, m;
int a, b, k; while(scanf("%d %d",&n,&m)!=EOF)
{
for(int i = 0; i < m; i++)
{
scanf("%d %d %d",&a,&b,&k);
ed[i].x=a,ed[i].y=b,ed[i].v=k;
}
Krusal(n,m);
}
return 0;
}
最新文章
- Xamarin技术文档------VS多平台开发
- ios webview自适应实际内容高度4种方法
- [转]搬瓦工换机房换ip之后不能连外网
- 理解assign,copy,retain变strong
- zabbix_agent安装(Centos+Ubuntu)
- 学习日记day9: PC端页面流程优化
- H264中的SPS、PPS提取与作用
- 目标识别:Bag-of-words表示图像
- 【USACO 2.3.2】奶牛家谱
- Android自定义工具类获取按钮并绑定事件(利用暴力反射和注解)
- Windows 2008 R2安装.NET Framework 4提示灾难性故障解决方法
- join on用法
- [bzoj省选十连测推广赛2]T2七彩树
- 电脑移动后WIFI连接失败解决方法
- Foundations of Game Engine Development Volume 1 Mathematics (Eric Lengyel 著)
- 编程一小时 code.org [六一关注]
- 团队第二次 # scrum meeting
- springboot与缓存(redis,或者caffeine,guava)
- web安全之攻击
- ci框架多语言切换