LightOJ - 1041 Road Construction(最小生成树)

题目大意：给你N条边，看能否形成最小生成树，如果存在，输出值，不存在，另外输出

解题思路：模版题

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <map>
#include <string>
#include <iostream>
using namespace std;
const int MAXNODE = 1010;
const int MAXEDGE = 1000010;
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];
map<string,int> Map;

//初始化并查集和最小生成树的边
void init() {
    Map.clear();
    n = 0;
    cin >> m;
    string a, b;
    int val;
    for (int i = 0; i < m; i++) {
        cin >> a >> b >> val;
        if (!Map[a]) Map[a] = ++n;
        if (!Map[b]) Map[b] = ++n;
        E[i] = Edge(Map[a], Map[b], val);
    }

    for (int i = 1; i <= n; i++) {
        f[i] = i;
        G[i].clear();
    }
}

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

bool cmp(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);
    }
}

//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, cmp);

    Type Sum = 0;
    int num = 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;
            Sum += 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]);
            num++;
        }
    }
    if(num != n - 1)
        cout << "Case " << cas++ << ": Impossible" << endl;
    else 
        cout << "Case " << cas++ << ": " << Sum << endl;
}

int main() {
    int test;
    cin >> test;
    while (test--) {
        init();
        solve();
    }
    return 0;
}