POJ 1861 & ZOJ 1542 Network(最小生成树之Krusal)

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.

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.

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.

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;
}
时间: 2024-11-20 12:26:13

POJ 1861 &amp; ZOJ 1542 Network(最小生成树之Krusal)的相关文章

POJ 1861 &amp; ZOJ 1542 Network(最小生成树之Krusal)

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 c

zoj 1542 Network

#include <stdio.h> #include <string.h> #include <iostream> #include <algorithm> #include <vector> #include <queue> #include <set> #include <map> #include <string> #include <math.h> #include <s

ZOJ 1542 POJ 1861 Network 网络 最小生成树,求最长边,Kruskal算法

题目连接:ZOJ 1542 POJ 1861 Network 网络 Network Time Limit: 2 Seconds      Memory Limit: 65536 KB      Special Judge 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, t

poj 1861 Network

Network Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 13260   Accepted: 5119   Special Judge 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 c

poj 1861 Network 解题报告

Network Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 16171   Accepted: 6417   Special Judge 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 c

poj 1861(最小生成树)

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

POJ 2349 Arctic Network 最小生成树题解

本题也是使用Prime和Kruskal都可以的最小生成树的题解. 本题一点新意就是:需要除去最大的S-1个距离,因为可以使用卫星覆盖这些距离. 技巧:建图建有向图,速度快点,不用计算两边. 这里使用Prime,因为是稠密图. #include <stdio.h> #include <math.h> #include <stdlib.h> #include <string.h> #include <limits.h> #include <al

poj 1861 Network (kruskal)

Network Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 13633   Accepted: 5288   Special Judge 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 c

POJ 1861 Kruskal

Network Time Limit: 1000MS  Memory Limit: 30000K Total Submissions: 13456  Accepted: 5209  Special Judge 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 comp