1489. 找到最小生成树里的关键边和伪关键边

思路

基础的kruskal基础上区分两种边，首先求一下MST的Mcost：

  • 关键边，删除该边，再求MST，若cost > Mcost (或者不连通，也满足大于)则该边为关键边
  • 伪关键边，非关键边有可能是伪关键边，可以先将该边加入MST，然后继续kruskal， 若cost == Mcost吗，则为非关键边，否则“啥也不是边”

AC code

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class Solution {
public:
    const static int inf = 0x3f3f3f3f;
    vector<int> p;

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

    int kruskal(int n, vector<vector<int>>& edges, vector<int>& del){
        p.resize(n);
        for(int i = 0; i < n; i++) p[i] = i;
        int cnt = 0;
        int cost = 0;
        for(auto e : edges){
            if(e == del) continue; 
            // printVector1d(e);
            int a = e[0], b = e[1], w = e[2];
            int pa = find(a), pb = find(b);
            if(pa != pb){
                p[pa] = pb;
                cnt++;
                cost += w;
            }
        }
        return cnt == n - 1 ? cost : inf;
    }

    int rekruskal(int n, vector<vector<int>>& edges, vector<int>& del){
        p.resize(n);
        for(int i = 0; i < n; i++) p[i] = i;
        p[del[0]] = del[1];
        int cnt = 1;
        int cost = del[2];
        for(auto e : edges){
            if(e == del) continue; 
            int a = e[0], b = e[1], w = e[2];
            int pa = find(a), pb = find(b);
            if(pa != pb){
                p[pa] = pb;
                cnt++;
                cost += w;
            }
        }
        return cnt == n - 1 ? cost : inf;
    }

    vector<vector<int>> findCriticalAndPseudoCriticalEdges(int n, vector<vector<int>>& edges) {
        int edgen = edges.size();
        for(int i = 0; i < edgen; i++) edges[i].push_back(i);
        sort(edges.begin(), edges.end(), [&](const vector<int>& a, const vector<int>& b){
            return a.at(2) < b.at(2);
        });
        vector<int> del;
        int minCost = kruskal(n, edges, del);
        vector<vector<int>> res(2);
        for(auto e : edges){
            del = e;
            int cost = kruskal(n, edges, del);
            //cout << i << " " << cost << endl;
            if(cost > minCost) res[0].push_back(e[3]);
            else if(rekruskal(n, edges, del) == minCost) res[1].push_back(e[3]);
        }
        return res;
    }
};