LightOj-1029-Civil and Evil Engineer

---

题目传送门

---

题意:给定一个无向图，求他的最小生成树与最大生成树的平均值。

思路:用Kruskal正着求最小生成树，再倒着求最大生成树。

#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;}