c_cpp 图表的DFS

Posted 2021-05-14

#include<bits/stdc++.h>
using namespace std;

// #Graphs #BasicProblem
//	https://www.geeksforgeeks.org/depth-first-search-or-dfs-for-a-graph/
// Given graph is a strongly connected directed graph
// (there is atleast 1 path from each vertex to another vertex)

// Testcase
// 1
// 4
// 6
// 0 2
// 0 1
// 1 2
// 2 0
// 2 3
// 3 3
// 2
// Ans: 2 0 1 3

void DFS(vector< vector<int> > AdjL,vector<bool> &visited,int start){
    cout<<start<<" ";
    visited[start]=true;
    for(int i=0;i<AdjL[start].size();i++){
        // loop through the current vertex's AdjL and recursively DFS if vertex is not visited
        if(visited[AdjL[start][i]]==false){
            DFS(AdjL,visited,AdjL[start][i]);
        }
    }
}

int main(){
    freopen("ip.txt","r",stdin);
    int t;
    cin>>t;
    while(t--){
      	int v;
    	cin>>v;
    	vector< vector<int> > AdjL(v);
    	int e;
    	cin>>e;
    	int i=0;
    	while(i<e){
    	    int a,b;
    	    cin>>a>>b;
    	    AdjL[a].push_back(b);
    	    i++;
    	}
    	vector<bool> visited(v,false);
    	int start;
    	// cout<<"Enter starting vertex"<<endl;
    	cin>>start;
    	cout<<"DFS: ";
    	DFS(AdjL,visited,start);
    	cout<<endl;
//======================(Iterative DFS)==========================
// https://leetcode.com/explore/learn/card/queue-stack/232/practical-application-stack/1385/
        for(int i=0;i<v;i++){
            visited[i]=false; // reset visited
        }
        stack<int> st; // contains id's of vertices
        st.push(start);
        visited[start]=true;
        cout<<"Iterative DFS: "<<endl;
        while(!st.empty()){
            int currTop=st.top(); // store top in temp var
            st.pop(); // and pop the current stack
            cout<<currTop<<" ";
            for(int i=AdjL[currTop].size()-1;i>=0;i--){ //for left to right
                if(!visited[AdjL[currTop][i]]){ // as left element is added last
                    st.push(AdjL[currTop][i]); // it comes out first
                    // cout<<currTop<<" ";
                    visited[AdjL[currTop][i]]=true;
                }
            }
        }
        cout<<endl;
    }
    return 0;
}