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;

}