Network

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 61 2 11 3 11 4 22 3 13 4 12 4 1

Sample Output

141 21 32 33 4

题解：样例假的吧。要求每一个最小生成树中长度最大的一条在所有生成树中最小。用prim似乎不好办，因为每次都选当前最短的一条，但是如果我不选这条也有可能得到最小生成树，并且最长的一条要小，所以要用克鲁斯卡尔，排序后，每次选取最小的一条，肯定是最优解。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int INF = 0x3fffffff;struct Node{int from;int to;int cost;bool operator< (Node t) const{return cost < t.cost;}};Node e[300005];int d[1003];int pre[1003];int res[1003];int maxlen;int t;int ans;int find(int x){return x == pre[x] ? x : pre[x] = find(pre[x]);}void kruskal(int n,int k){int x = 0;t = 0;ans = 0;    for(int i = 1;i < n;i++)    {    for(int j = x;j < k;j++)    {    int x = find(e[j].from);    int y = find(e[j].to);    if(x == y)    {    continue;}pre[x] = y;res[t++] = j;ans = max(ans,e[j].cost);x = j + 1;break;}}}int main(){int n,m;while(scanf("%d%d",&n,&m) != EOF){int u,v,c;int k = 0;for(int i = 1;i <= n;i++){pre[i] = i;}for(int i = 0;i < m;i++){scanf("%d%d%d",&u,&v,&c);e[k].from = u;e[k].to = v;e[k++].cost = c;e[k].from = v;e[k].to = u;e[k++].cost = c;}sort(e,e + k);kruskal(n,k);printf("%d\n%d\n",ans,n - 1);for(int i = 0;i < n - 1;i++){printf("%d %d\n",e[res[i]].from,e[res[i]].to);}}return 0;}