POJ 1861 Network (Kruskal求MST模板題)

 

Network Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 14103   Accepted: 5528   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

題目鏈接：poj.org/problem?id=1861

題目大意：n個點，m條線，每條線有個權值，現在要求最長的路最短且讓各個點都連通，求最短的最長路，邊個數和對應邊

題目分析：樣例有問題，應該是
1
4
1 2
1 3
3 4
裸的Kruskal注意這裡要求最長路最短，而Kruskal正好是對權值從小到大排序後的貪心算法

#include 
#include 
#include 
using namespace std;
int const MAX = 15005;
int fa[MAX];
int n, m, ma, num;
int re1[MAX], re2[MAX]; 

struct Edge
{
    int u, v, w;
}e[MAX];

bool cmp(Edge a, Edge b)
{
    return a.w < b.w;
}

void UF_set()
{
    for(int i = 0; i < MAX; i++)
        fa[i] = i;
}

int Find(int x)
{
    return x == fa[x] ? x : fa[x] = Find(fa[x]);
}

void Union(int a, int b)
{
    int r1 = Find(a);
    int r2 = Find(b);
    if(r1 != r2)
        fa[r2] = r1;
}

void Kruskal()
{
    UF_set();
    for(int i = 0; i < m; i++)
    {
        int u = e[i].u;
        int v = e[i].v;
        if(Find(u) != Find(v))
        {
            re1[num] = u;
            re2[num] = v;
            Union(u, v);
            ma = max(ma, e[i].w);
            num ++;
        }
        if(num >= n - 1)
            break;
    }
}

int main()
{  
    ma = 0;
    num = 0;
    scanf("%d %d", &n, &m);
    for(int i = 0; i < m; i++)
        scanf("%d %d %d", &e[i].u, &e[i].v, &e[i].w);
    sort(e, e + m, cmp);
    Kruskal();
    printf("%d\n%d\n", ma, num);
    for(int i = 0; i < num; i++)
        printf("%d %d\n", re1[i], re2[i]);
}