MST边检查(MST Edge Check) 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;

struct Edge {
    int u, v, w, 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;
    }
    sort(edges.begin(), edges.end());
    vector<int> p(n + 1);
    vector<char> ans(m, 0);
    iota(p.begin(), p.end(), 0);
    auto find = [&](int x) {
        while (p[x] != x) x = p[x] = p[p[x]];
        return x;
    };
    for (int i = 0; i < m; ) {
        int j = i;
        while (j < m && edges[j].w == edges[i].w) ++j;
        for (int k = i; k < j; ++k) {
            if (find(edges[k].u) != find(edges[k].v)) ans[edges[k].id] = 1;
        }
        for (int k = i; k < j; ++k) {
            int ru = find(edges[k].u), rv = find(edges[k].v);
            if (ru != rv) p[ru] = rv;
        }
        i = j;
    }
    for (int i = 0; i < m; ++i) cout << (ans[i] ? "YES\n" : "NO\n");
    return 0;
}