深度优先搜索 dfs Matt's Graph Book

中文题意：就是一个图，去掉一条边后问你连不连通

Matt's Graph Book

Attempted by: 1659

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Accuracy: 52%

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Maximum Score: 20

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4 Votes

Tag(s):

 

Algorithms, Easy

PROBLEM

EDITORIAL

MY SUBMISSIONS

ANALYTICS

Matt loves Graph Theory. He recently enrolled into the Algorithms course and started liking it too. His teacher understood his love for graphs and decided to give him a problem.

She gives him a Tree and asks him to write a program to find if it's connected or not. She realizes that Matt would do this in no time. So, instead she gives him a Graph and then removes a vertex from it and now asks him to find if the resultant graph is connected or not.

Since he is new to programming and algorithms, you have to help him solve this problem.

Input

First line contains a single integer N denoting the number of vertices in the graph. Second line contains a single integer k denoting the number of edges in the graph. k lines follow each containing two space separated integers a and b denoting the edge between the vertices a and b. Then, the last line contains a single integer x which denotes the vertex that is removed from the graph.

Output

"Connected" (without quotes) if the resultant graph is connected. "Not Connected" (without quotes), otherwise.

Constraints

1 ≤ N ≤ 10⁵
1 ≤ k ≤ 2×10⁵
0 ≤ a,b ≤ 10⁵
0 ≤ x ≤ 10⁵

SAMPLE INPUT

 

433 00 11 22

SAMPLE OUTPUT

 

Connected

Explanation

Number of vertices: 4 
Number of edges: 3

It is still Connected.

Time Limit:2.0 sec(s) for each input file.

Memory Limit:256 MB

Source Limit:1024 KB

#include <bits/stdc++.h>#include <vector>#define mod 1000000007using namespace std;const int mn=200005;vector <int>adj[200005];bool visited[200005]={0};void dfs(int s){    visited[s]=1;    for(int i=0;i<adj[s].size();++i)    {        if(!visited[adj[s][i]])            dfs(adj[s][i]);    }}int main(){    int nodes,edges;    cin>>nodes>>edges;    int u,v;    for(int i=0;i<edges;++i)    {        cin>>u>>v;        adj[u].push_back(v);        adj[v].push_back(u);    }    int x;    cin>>x;    adj[x].clear();    visited[x]=1;    if(x!=0)dfs(0);    else dfs(1);    bool ok=true;    for(int i=0;i<nodes;++i)    {        if(!visited[i]){            ok=false;            break;        }    }    if(!ok)        cout<<"Not Connected";    else        cout<<"Connected";    return 0;}