LightOj-1029-Civil and Evil Engineer

---

題目傳送門

---

題意:給定一個無向圖，求他的最小生成樹與最大生成樹的平均值。

思路:用Kruskal正着求最小生成樹，再倒着求最大生成樹。

#include <bits/stdc++.h>

using namespace std;

struct node{
    int x, y, w;
}edge[30000];
int cnt;
int n;
int fa[15000];
int cmp1(node a, node b)
{
    return a.w < b.w;
}
int cmp2(node a, node b)
{
    return a.w > b.w;
}

int find(int x)
{
    if (fa[x]!=x)
        fa[x] = find(fa[x]);
    return fa[x];
}

int Kruskal()
{
    int sum = 0;
    for (int i = 0; i <= n; i++)
    {
        fa[i] = i;
    }
    for (int i = 0; i < cnt; i++)
    {
        int a = find(edge[i].x);
        int b = find(edge[i].y);
        if (a!=b)
        {
            fa[a] = b;
            sum+=edge[i].w;
        }
    }
    return sum;
}

int main(void)
{
    int T, cas=1;
    scanf("%d", &T);
    while (T--)
    {
        cnt = 0;
        scanf("%d", &n);
        int u, v, w;
        while (scanf("%d %d %d", &u, &v, &w))
        {
            if (u+v+w==0)
                break;
            edge[cnt].x = u;
            edge[cnt].y = v;
            edge[cnt++].w = w;
            edge[cnt].x = v;
            edge[cnt].y = u;
            edge[cnt++].w = w;
        }
        int ans = 0;
        sort(edge, edge+cnt, cmp1);
        ans += Kruskal();
        sort(edge, edge+cnt, cmp2);
        ans += Kruskal();
        if (ans&1)
            printf("Case %d: %d/2\n", cas++, ans);
        else
            printf("Case %d: %d\n", cas++, ans/2);
    }
    return 0;
}