DMOJ · nccc5j5s3

Directed Graph Connectivity

This C++ solution uses graph traversal for DMOJ nccc5j5s3 Directed Graph Connectivity. Read the reasoning, inspect the code, or try your own test case below.

nccc5j5s3Graphs & treesGraph traversalC++38 lines
Solution076of 248
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Approach

Graph traversal

Directed Graph Connectivity asks, for every edge removed in turn, whether 1 can still reach N; the code performs that repeated reachability test.

Graphs & trees

Problem and code

Useful links.

Written by benbenyaojifen. Try the problem first, then compare your approach with the code.

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Implementation

directed_graph_connectivity.cpp

C++

    #include <bits/stdc++.h>
     
    using namespace std;
     
    void dfs(int n, int indx, pair<int, int> bad, vector<bool> &visited, vector<vector<int>> &adj){
        if(visited[indx]) return;
        visited[indx] = true;
        if(indx == n) return;
        for(int c : adj[indx]){
            if(!(indx == bad.first && c == bad.second)) dfs(n, c, bad, visited, adj);
        }
    }
    int main(){
        ios::sync_with_stdio(0);
        cin.tie(0);
        int n, m;
        cin >> n >> m;
        vector<vector<int>> adj(n + 1);
        unordered_map<int, pair<int, int>> day;
        for(int i = 0; i <m; i++){
            int x, y;
            cin >> x >> y;
            x--; y--;
            adj[x].push_back(y);
            day[i] = make_pair(x, y);
        }
        for(int i = 0; i < m; i++){
            pair bad = day[i];
            vector<bool> visited(n, false);
            dfs(n - 1, 0, bad, visited, adj);
            if(visited[n - 1]){
                cout << "YES\n";
            } else {
                cout << "NO\n";
            }
        }
        return 0;
    }
        

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