洛谷 P1656 炸鐵路

題意：n個點，m條邊，問有哪條邊是去掉之後，會造成之前連通的點不再連通的？n <= 150, m <= 5000.

思路：連通演算法有dfs+bool陣列記錄，或者dsu，感覺dsu更方便。m * n 不超過1e6，直接暴力。

class DisjointSet{
public:
    DisjointSet(int sz): sz_(sz){
        set_size_.assign(sz_, 1);
        fa_.resize(sz_);
        iota(fa_.begin(), fa_.end(), 0);
    }

    int findSet(int x){ return fa_[x] == x ? x : fa_[x] = findSet(fa_[x]); }

    int getSetSize(int x){ return set_size_[findSet(x)]; }

    bool unionSet(int x, int y){
        x = findSet(x);
        y = findSet(y);
        if (x == y){
            return false;
        }
        fa_[x] = y;
        set_size_[y] += set_size_[x];
        return true;
    }

private:
    int sz_;
    vector<int> fa_;
    vector<int> set_size_;

};


void solve(){
    int n, m;
    cin >> n >> m;

    DisjointSet dsu(n + 1);
    vector<vector<int>> al(n + 1);
    vector<pair<int, int>> edges;
    for (int i = 1; i <= m; ++i){
        int u, v;
        cin >> u >> v;
        dsu.unionSet(u, v);
        if (u > v){
            swap(u, v);
        }
        edges.emplace_back(u, v);
    }

    vector<int> pos;
    for (int i = 0; i < m; ++i){
        DisjointSet dsu2(n + 1);
        for (int j = 0; j < m; ++j){
            if (j != i){
                dsu2.unionSet(edges[j].first, edges[j].second);
            }
        }
        for (int k = 1; k <= n; ++k){
            if (dsu2.getSetSize(k) != dsu.getSetSize(k)){
                pos.emplace_back(i);
                break;
            }
        }
    }

    sort(pos.begin(), pos.end(), [&](int i, int j){
            return edges[i].first != edges[j].first ? edges[i].first < edges[j].first : edges[i].second < edges[j].second;
         });

    for (const int& index : pos){
        cout << edges[index].first << " " << edges[index].second << "\n";
    }

}

總結：使用下標陣列排序，可以避免再次儲存答案佔用更多記憶體。 unionSet裡面返回值一定要寫。