62 lines
1.3 KiB
C++
62 lines
1.3 KiB
C++
#include <iostream>
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#include <vector>
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using namespace std;
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int main(){
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//dp只能O(n3)，这里最好用bfs或者dfs
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int v, e;
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cin >> v >> e;
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//vector<vector<int>> adj(v, vector<int>());
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vector<vector<bool>> dp(v, vector<bool>(v, false));
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for(int i = 0; i < e; i++){
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int a, b;
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cin >> a >> b;
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//adj[a - 1].push_back(b - 1);
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//adj[b - 1].push_back(a - 1);
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dp[a - 1][b - 1] = true;
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}
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//考虑dp[i][j] : 两个点是否连通，采取插入k的方式解决
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for (int i = 0; i < v; i++) dp[i][i] = true; // 自己到自己连通
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for(int k = 0; k < v; k++){
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for(int i = 0; i < v; i++){
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for(int j = 0; j < v; j++){
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if(dp[i][j]) continue;
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dp[i][j] = dp[i][j] || (dp[i][k] & dp[k][j]);
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}
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}
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}
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//bfs
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for(int i = 0; i < v; i++) {
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vector<bool> visited(v, false);
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queue<int> q;
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q.push(i);
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visited[i] = true;
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int mx = i;
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while (!q.empty()) {
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int curr = q.front();
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q.pop();
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for (int next : adj[curr]) {
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if (!visited[next]) {
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visited[next] = true;
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q.push(next);
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mx = max(mx, next);
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}
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}
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}
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cout << mx + 1 << " ";
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}
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for(int i = 0; i < v; i++){
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int mx = -1;
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for(int j = 0; j < v; j++){
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if(dp[i][j] && mx < j) mx = j;
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}
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cout << mx + 1 << " ";
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}
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cout << endl;
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return 0;
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}
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