POJ-1861-NETWORK 解题报告

Network

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 16628		Accepted: 6597		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 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

好久没有更新这个博客，觉得自己还是不够努力，自己的acm学习之路的脚步走的很慢，希望尽快改变现状吧。这道题考察kruskal算法，留下个记录

题解：题目大意是有n个顶点，m条边，使图连通，使边权和最小，不构成回路

kruskal算法：按边权从小到到排序，顺序将边加入图中，若图中两点不连通则加入，否则就不考虑，（判断两点是否在图中中并查集，并查集的主要功能就是合并和查找的功能），最终就可以解答此题，找到最小的边权的连通图。

初始化并查集f[]数组（f中每个数等于下标值）find(x),

查找x的祖先，若x!=find(x)则find(find(x))，递归操作即可

合并操作，union(x,y) 若两个是不同的祖先则合并。

#include <stdio.h>
#include <iostream>
#include <cstdio>
#include <algorithm>

using namespace std;

const int maxn = 1e6+7;

int father[maxn];

int n, m, c = 1;
struct _edge{
    int x, y, d;
}edge[maxn], ans[maxn];

bool cmp(_edge a, _edge b)
{
    return a.d < b.d;
}

int fi(int x)
{
    return x == father[x] ? x : father[x] = fi(father[x]);
}

void union1(int x, int y)
{
    int p1 = fi(x), p2 = fi(y);
    if (p1 == p2) return;
    father[p1] = p2;
}

int check(int x, int y)
{
    int p = fi(x), q = fi(y);
    if (p == q)
        return 1;
    return 0;
}

void init()
{
    for (int i=0; i<=n; i++)
        father[i] = i;
}

void kruskal()
{
    for (int i=1; i<=m; i++)
    {
        int x = edge[i].x;
        int y = edge[i].y;
        if (fi(x) != fi(y)) {
            union1(x, y);
            ans[c].x = x;
            ans[c].y = y;
            ans[c].d = edge[i].d;
            c++;
        }
    }
}

int main()
{
    int x, y;
    scanf("%d%d", &n, &m);
    init();
    for (int i=1; i<=m; i++)
    {
        scanf("%d%d%d", &edge[i].x, &edge[i].y, &edge[i].d);
    }
    sort(edge+1, edge+m+1, cmp);
    kruskal();
    printf("%d\n", ans[c-1].d);
    printf("%d\n", c-1);
    for (int i=1; i<c; i++) {
        printf("%d %d\n", ans[i].x, ans[i].y);
    }
    return 0;
}