CODEWITHSUNDEEP
c++algorithm
prashantgpt91
prashantgpt91

Reputation: 1795

Connected Components in an undirected graph

i correctly coded this question and get accepted in 0.39 sec using bfs algorithm

here's my code

#include<iostream>
#include<vector>
#include<string.h>
#include<cstdio>
#include<queue>
using namespace std;

int main()
{
queue<int>q;
bool visited[100000];
int t,i,j,x,y;
int n,e;
cin>>t;
while(t--)
{
scanf("%d%d",&n,&e);
vector< vector<int> >G(n);
memset(visited,0,sizeof(visited));

for(i=0;i<e;i++)
{
scanf("%d%d",&x,&y);
G[x].push_back(y);
G[y].push_back(x);
}

int ans=0;
for(i=0;i<n;i++)
{
if(!visited[i])
{
q.push(i);
visited[i]=1;

while(!q.empty())
{
int p=q.front();
q.pop();
for(j=0;j<G[p].size();j++)
{
if(!visited[G[p][j]])
{
visited[G[p][j]]=1;
q.push(G[p][j]);
}
}
}
ans++;
}
}
printf("%d\n",ans);
}
}

but then i checked another accepted code in 0.15 sec i don't understand what he has done. Some body please explain.

i simply understand that he did not used bfs or dfs algorithm. and why his method got accepted in lesser time.

here's is his code

#include <iostream>
#include <vector>
#include <stdio.h>

using namespace std;

int p[100001];


int find( int a)
{
int root = a;
while ( p[a] != a) {
    a = p[a];
}
while ( root != a) {

  int root2 = p[root];
  p[root] = a;
  root = root2;
}   

return a;

} 

int main()
  {
int t;

scanf("%d",&t);
while ( t-- ) {
    int n,m,a,b,q;

    scanf("%d%d",&n,&m);
    for ( int i = 0; i < n; i++) 
        p[i] = i;

    //int count = n;

    for ( int i = 0; i < m;i++) {

        scanf("%d%d",&a,&b);
        int root = find(a);
        int root2 = find(b);
        p[root2] = root;

    }
    int count = 0;
    for ( int i = 0; i < n;i++) {
        if ( p[i] == i)
          count++;

    }
    printf("%d\n",count);
}
return 0;

}

Upvotes: 0

Views: 828

Answers (1)

Ivaylo Strandjev
Ivaylo Strandjev

Reputation: 71019

This code uses disjoint set forest. A very efficient data structure that is able to perform union-find. In essence the user has a set of sets and can perform two operations:

  • join two sets
  • check which set the element belongs to

The algorithm above also uses one important heuristics - path compression. Usually disjoint set is implement using one more heuristics - union by rank but the author decided not to use it.

Upvotes: 2

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