Question

A tennis player N players have plan to match each other total M pairs, one match per day and no two tennis player match more than one During that time, someone want to organize a tug of war battle,by devide a tennis player into two team, but to unite playing tug of war in each team must have player who never do tennis competition before and this is more important than number of player in both team Write a program that receive a Competition Schedule then calculate day at the lastest that still can divide a tennis player for playing tug of war after tennis competition Input First line specify 2 integer N and M (2 N< 100,000; 1< M<-100,000) After that M line will spcify competition schedule ie in line 1+I will specify 2 integer A and B for telling that in day 1 Platyer A will match with player B (1<-A?N; 1?B<=N) and garuntee that any pair of player won't play more than 1 day Output 1 line represent day that can divide tennnis player into two team for playing tug of war, player in the same team will never competition tennis before Example1 Input Output 12 2 3 3 1 Example2 Input 5 7 12 2 3 Output 6 5 2 5 1

Answer 1

#include<bits/stdc++.h>
#define lli long long int
#define pb push_back
#define INF 1e18
#define N 100010

using namespace std;

/*

Main idea behind my implementation is as follows:

We can't have tug of war if we can't find two set of players such that no one has played againest each other
before

To have this , we have a specific algorithm called bipartiteness
If the graph obtained till the match on that day don't violate then we add that day

Once we have found it violates bipartiteness on a day , it will violate one the next days as well.

bipartiteness can be checked by performing a dfs which has a time complexity of O(V+E)

*/

vector<lli> adj[N];
bool visited[N];
lli color[N];

//dfs for checking bipartiteness
bool dfs(lli s,lli p)
{
   visited[s] = 1;
   lli i,v;
   color[s] = 1-color[p];//alternate color in neighbours
   for(i=0;i<adj[s].size();i++)
   {
       v = adj[s][i];
       if(!visited[v])
       {
           if(!dfs(v,s))
               return 0;
       }
       else
       {
           if(color[v] == color[s])//if two neighbours have same color then bipartiteness is violated , hence return false
               return 0;
       }
   }
   return 1;//return true if till now bipartiteness is followed
}

int main()
{
   lli n,m,i,j,k,u,v,ans;
   bool valid = 1;//to check if tog of war is possible
   cin>>n>>m;
  
  
   ans = 0;
   for(i=1;i<=m;i++)
   {
       cin>>u>>v;
       adj[u].pb(v); adj[v].pb(u);
      
       //if already violated then no pointing in checking further
       if(!valid)
           continue;
      
       memset(visited,0,sizeof visited);
       color[0] = 0;//0 is an aribrary node which is virtually taken as parent of roots for performing dfs
      
       for(j=1;j<=n;j++)
       {
           if(visited[j])
               continue;
           valid = dfs(j,0);
           if(!valid)
               break;
       }
       if(valid)
           ans++;
   }
  
  
   cout<<ans<<endl;
  
}

CA File Edit Search View Project Execute Tools AStyle WindowHelp PROSisth.cpp-Executing)- Dev-C+5.11 sth.cpp #include< bits/stdc++.h> #define 111 long long int 3 #define pb push back #define INF 1e1 CAUsers Luc 5 #define N 10001 8 using namespaco 9 12 13 Main idea behi rocess exited after 6.747 seconds with return value θ ress any key to continue 15 We can't have t 16 before 17 18 To have this 19 If the graph ob 20 21 Once we have fo , 23 bipartiteness c 24 25 26*/ 27 28 vectorくlli>ad 29 bool visited IN] 30 ll ; 31 32 IIdfs for checking bipartiteness i color[N] Line: Col: 10 Sel: 0 Lines: 94 1826 Insert Done in 0.016 seconds

CA File Edit Search View Project Execute Tools AStyle WindowHelp PROSisth.cpp-Executing)- Dev-C+5.11 sth.cpp #include< bits/stdc++.h> #define 111 long long int 3 #define pb push-back #define INF 1e18 5 #define N 100010 CAUsersilu 8 using namespace std; 9 12 13 Main idea behind my implementat 15 We can't have tug of war if we c 16 before 17 18 To have this , we have a specif 19 If the graph obtained till the 20 21 Once we have found it violatesb Process exited after 22.26 seconds with return value θ ress any key to continue .. . 23 bipartiteness can be checked by 24 25 26*/ 27 28 29 bool visited IN]; 30 lli color [N]; 31 vector<lli> adj[N]i O Type here to search 11:25 AM A<5/27/2018