洛谷 P1656 炸铁路

题意：n个点，m条边，问有哪条边是去掉之后，会造成之前连通的点不再连通的？n <= 150, m <= 5000.

思路：连通算法有dfs+bool数组记录，或者dsu，感觉dsu更方便。m * n 不超过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 fa_;
vector set_size_;

};

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

DisjointSet dsu(n + 1);
vector<vector> 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 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";
}

}

总结：使用下标数组排序，可以避免再次存储答案占用更多内存。 unionSet里面返回值一定要写。