LightOJ - 1029 Civil and Evil Engineer(最大/小生成树)

题目大意：问最小生成树和最大生成树的平均值

解题思路：模版题

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;
const int MAXNODE = 110;
const int MAXEDGE = 12010;
typedef int Type;

struct Edge{
    int u, v;
    Type d;
    Edge() {}
    Edge(int u, int v, Type d): u(u), v(v), d(d) {}
}E[MAXEDGE];

int n, m, tot, cas = 1;
int f[MAXNODE];
Type maxcost[MAXNODE][MAXNODE];
vector<Edge> G[MAXNODE];

//初始化并查集和最小生成树的边
void init() {
    scanf("%d", &n);
    for (int i = 0; i <= n; i++) {
        f[i] = i;
        G[i].clear();
    }
    m = 0;
    while (1) {
        scanf("%d%d%d", &E[m].u, &E[m].v, &E[m].d);
        if(E[m].u == 0 && E[m].v == 0 && E[m].d == 0) break;
        m++;
    }
    m++;
}

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

bool cmp1(const Edge &a, const Edge &b) {
    return a.d < b.d;
}

bool cmp2(const Edge &a, const Edge &b) {
    return a.d > b.d;
}

//dfs找路径最大值,maxcost[i][j]维护的是树上的i到j点的路径上，最长的那条边的权值 
void dfs(int s, int u, Type Max, int fa) {
    maxcost[s][u] = max(maxcost[s][u], Max);
    for (int i = 0; i < G[u].size(); i++) {
        int v = G[u][i].v;
        if (v == fa) continue;
        double tmp = max(Max, G[u][i].d);
        dfs(s, v, tmp, u);
    }
}

int gcd(int a, int b) {
    if (a % b == 0) return b;
    return gcd(b, a % b);
}

//Kruskal找到最小生成树，并将最小生成树记录下来，以便后面用来求两点之间的最长边
void solve() {
    sort(E, E + m, cmp1);

    Type Min = 0;
    for (int i = 0; i < m; i++) {
        int fx = find(E[i].u);
        int fy = find(E[i].v);
        if (fx != fy) {
            f[fx] = fy;
            Min += E[i].d;
            G[E[i].u].push_back(E[i]);
            swap(E[i].u, E[i].v);
            G[E[i].u].push_back(E[i]);
        }
    }
    for (int i = 0; i <= n; i++)
        f[i] = i;
    sort(E, E + m, cmp2);

    Type Max = 0;
    for (int i = 0; i < m; i++) {
        int fx = find(E[i].u);
        int fy = find(E[i].v);
        if (fx != fy) {
            f[fx] = fy;
            Max += E[i].d;
            G[E[i].u].push_back(E[i]);
            swap(E[i].u, E[i].v);
            G[E[i].u].push_back(E[i]);
        }
    }
    if ((Max + Min) % 2 == 0)
        printf("Case %d: %d\n", cas++, (Max + Min) / 2);
    else 
        printf("Case %d: %d/2\n", cas++, Max + Min);
}

int main() {
    int test;
    scanf("%d", &test);
    while (test--) {
        init();
        solve();
    }
    return 0;
}