用c++解决1160. Network Time limit: 1.0 second Memory limit: 64 MB 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 contains two integer: 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.

这道题目是一个最小生成树的问题，可以使用Kruskal或Prim算法解决。

我们可以使用一个结构体来存储每个连接的信息，包括连接的两个节点以及连接的距离。然后将所有的连接按照距离从小到大排序，逐个加入最小生成树中，直到所有的节点都被连接。

在加入连接时，需要判断连接的两个节点是否在同一个连通块中，如果在，则不需要加入该连接，否则需要将两个连通块合并。

下面是使用Kruskal算法的c++代码：

#include <iostream>
#include <algorithm>
using namespace std;

const int MAXN = 1000+10;
const int MAXM = 15000+10;

struct Edge {
    int u, v, w;
    bool operator<(const Edge& e) const {
        return w < e.w;
    }
} e[MAXM];

int p[MAXN];

int find(int x) {
    return p[x] == x ? x : p[x] = find(p[x]);
}

int main() {
    int n, m;
    cin >> n >> m;
    for (int i = 0; i < m; i++) {
        cin >> e[i].u >> e[i].v >> e[i].w;
    }
    sort(e, e+m);
    for (int i = 1; i <= n; i++) {
        p[i] = i;
    }
    int ans = 0;
    for (int i = 0; i < m; i++) {
        int u = e[i].u, v = e[i].v, w = e[i].w;
        int pu = find(u), pv = find(v);
        if (pu != pv) {
            p[pu] = pv;
            ans = max(ans, w);
        }
    }
    cout << ans << endl;
    int cnt = 0;
    for (int i = 1; i <= n; i++) {
        if (p[i] == i) {
            cnt++;
        }
    }
    cout << cnt << endl;
    for (int i = 1; i <= n; i++) {
        if (p[i] == i) {
            cout << i << " ";
            for (int j = i+1; j <= n; j++) {
                if (p[j] == i) {
                    cout << j << endl;
                }
            }
        }
    }
    return 0;
}

在代码中，我们使用了一个数组p来存储每个节点所在的连通块的根节点，初始时每个节点自成一个连通块，即p[i] = i。

然后我们将所有的连接按照距离从小到大排序，依次加入最小生成树中。在加入连接时，我们判断连接的两个节点所在的连通块是否相同，如果不同，则将两个连通块合并，并且更新最大距离。

最后，我们统计连通块的个数，并输出每个连通块的节点。

时间复杂度为O(m log m)，其中m为连接的个数。