POJ-2485 Highways 最小生成树

一道简单的最小生成数，求使得所有的路连通的最小总路程代价中的最长的子路的长度。

代码如下：

#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <algorithm>
using namespace std;

int N, pos, set[505];

struct Node
{
    int x, y, dist;
    bool operator < (Node t) const 
    {
        return dist < t.dist;
    }
}e[250005];

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

void merge(int a, int b)
{
    set[a] = b;
}

int main()
{
    int T, c, cnt;
    scanf("%d", &T);
    while (T--) {
        pos = cnt = 0;
        scanf("%d", &N);
        for (int i = 1; i <= N; ++i) {
            set[i] = i;
            for (int j = 1; j <= N; ++j) {
                scanf("%d", &c);
                ++pos;
                e[pos].x = i, e[pos].y = j, e[pos].dist = c;
            }
        }  
        sort(e+1, e+1+pos);
        for (int i = 1; i <= pos; ++i) {
            int a = find(e[i].x), b = find(e[i].y);
            if (a != b) {
                merge(a, b);
                ++cnt;
                if (cnt == N-1) {
                    printf("%d\n", e[i].dist);
                    break;
                }
            }
        }
    }
    return 0;
}