最小生成树花费(Minimum Spanning Tree Cost) 题解

思路分析

按题目要求使用对应数据结构与算法模拟/递推求解，注意数据范围边界与时间复杂度。

参考代码

#include <iostream>
#include <vector>
#include <numeric>
#include <algorithm>
using namespace std;

const int LOG = 19;

struct Edge {
    int u, v;
    long long w;
    int id;
    bool operator<(const Edge& o) const { return w < o.w; }
};

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n, m;
    if (!(cin >> n >> m)) return 0;
    vector<Edge> edges(m);
    for (int i = 0; i < m; ++i) {
        cin >> edges[i].u >> edges[i].v >> edges[i].w;
        edges[i].id = i;
    }
    vector<Edge> sorted = edges;
    sort(sorted.begin(), sorted.end());
    vector<int> p(n + 1);
    iota(p.begin(), p.end(), 0);
    auto find = [&](int x) {
        while (p[x] != x) x = p[x] = p[p[x]];
        return x;
    };
    vector<vector<pair<int, long long>>> tree(n + 1);
    long long mstW = 0;
    for (const auto& e : sorted) {
        int ru = find(e.u), rv = find(e.v);
        if (ru != rv) {
            p[ru] = rv;
            mstW += e.w;
            tree[e.u].push_back({e.v, e.w});
            tree[e.v].push_back({e.u, e.w});
        }
    }
    vector<int> dep(n + 1, 0), q;
    vector<vector<int>> up(LOG, vector<int>(n + 1, 0));
    vector<vector<long long>> mx(LOG, vector<long long>(n + 1, 0));
    q.reserve(n);
    q.push_back(1);
    dep[1] = 1;
    for (int idx = 0; idx < (int)q.size(); ++idx) {
        int u = q[idx];
        for (auto [v, w] : tree[u]) {
            if (!dep[v]) {
                dep[v] = dep[u] + 1;
                up[0][v] = u;
                mx[0][v] = w;
                q.push_back(v);
            }
        }
    }
    for (int k = 1; k < LOG; ++k) {
        for (int i = 1; i <= n; ++i) {
            int mid = up[k - 1][i];
            up[k][i] = up[k - 1][mid];
            mx[k][i] = max(mx[k - 1][i], mx[k - 1][mid]);
        }
    }
    auto maxOnPath = [&](int u, int v) {
        long long res = 0;
        if (dep[u] < dep[v]) swap(u, v);
        int diff = dep[u] - dep[v];
        for (int k = 0; k < LOG; ++k) {
            if ((diff >> k) & 1) {
                res = max(res, mx[k][u]);
                u = up[k][u];
            }
        }
        if (u == v) return res;
        for (int k = LOG - 1; k >= 0; --k) {
            if (up[k][u] != up[k][v]) {
                res = max({res, mx[k][u], mx[k][v]});
                u = up[k][u];
                v = up[k][v];
            }
        }
        return max({res, mx[0][u], mx[0][v]});
    };
    for (int i = 0; i < m; ++i) {
        long long best = maxOnPath(edges[i].u, edges[i].v);
        cout << (mstW + edges[i].w - best) << "\n";
    }
    return 0;
}