PTA 最小生成树的唯一性——次小生成树

先dfs一遍求一下连通块的个数，连通块为1的情况下求出最小生成树和次小生成树，判断两者是否相等

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
using namespace std;
const int maxn = 510;
const int INF = 0x3f3f3f3f;
int n, m;
vector<int> G[maxn];
struct Edge {
    bool flag;
    int u, v, cost;
    bool operator < (const Edge &e) const {
        return cost < e.cost;
    }
}edge[maxn*maxn];
bool vis[maxn];
void dfs(int u) {
    for (int i = 0; i < G[u].size(); i++) {
        int v = G[u][i];
        if (vis[v]) continue;
        vis[v] = true;
        dfs(v);
    }
}
vector<int> p[maxn];
int dis[maxn][maxn];
int fa[maxn];
int query(int x) { return fa[x] == x ? x : fa[x] = query(fa[x]); }
void mst() {
    for (int i = 1; i <= n; i++) p[i].push_back(i);
    for (int i = 1; i <= n; i++) fa[i] = i;
    sort(edge, edge+m);
    int cnt = 0;
    for (int i = 0; i < m && cnt < n-1; i++) {
        int u = edge[i].u, v = edge[i].v, cost = edge[i].cost;
        u = query(u), v = query(v);
        if (u == v) continue;
        for (int j = 0; j < p[u].size(); j++) {
            for (int k = 0; k < p[v].size(); k++) {
                int x = p[u][j], y = p[v][k];
                dis[x][y] = dis[y][x] = cost;
            }
        }
        for (int k = 0; k < p[v].size(); k++) p[u].push_back(p[v][k]);
        fa[v] = u;
        edge[i].flag = true;
        cnt++;
    }
}
int main() {
    cin >> n >> m;
    int u, v, cost;
    for (int i = 0; i < m; i++) {
        cin >> u >> v >> cost;
        G[u].push_back(v);
        G[v].push_back(u);
        edge[i].u = u, edge[i].v = v, edge[i].cost = cost, edge[i].flag = false;
    }
    int cnt = 0;
    memset(vis, false, sizeof(vis));
    for (int i = 1; i <= n; i++) {
        if (!vis[i]) {
            vis[i] = true;
            cnt++;
            dfs(i);
        }
    }

    if (cnt > 1) {
        cout << "No MST" << endl << cnt << endl;
    }
    else {
        mst();
        int sum = 0, ans = INF;
        for (int i = 0; i < m; i++) if (edge[i].flag) sum += edge[i].cost;
        for (int i = 0; i < m; i++) {
            if (edge[i].flag) continue;
            ans = min(ans, sum-dis[edge[i].u][edge[i].v]+edge[i].cost);
        }
        cout << sum << endl;
        if (ans != sum) cout << "Yes" << endl;
        else cout << "No" << endl;
    }
    return 0;
}

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