Solution for MIKE3

#include<cstdio>

int read(){
char c=getchar_unlocked();
 int n=0;
while(!(c>='0' && c<='9'))
c=getchar_unlocked();
while(c>='0' && c<='9'){
n=n*10 + (c-'0');
c=getchar_unlocked();
}
return n;
}
int stamp[25][20005]={{0}};//orignally storing the values of stamps j which are asked by the ith group in stamp[i][j]
int graph[25][25]={{0}};//this stores the groups which intersect with other groups
int after[25]={0};//sum of all the elements in the graph (coloumn --) which denotes the number of collision for that group
int possible[25]={0};//value is one if the group of stamp is selected

int summation[20005]={0};

int main(){
    int N,M;
    N=read();
    M=read();

    for(int i=1;i<=M;i++){
        int x;
        x=read();
        stamp[i][0]=x;
        for(int j=1;j<=x;j++){int y;
        y=read();
        stamp[i][y]=1;
        }
    }


    for(int j=1;j<=N;j++){
        int sum=0;
    for(int i=1;i<=M;i++){
        if(i==j)
        graph[i][j]=1;// summing up all the coloumns in the stamps matrix which denotes how many groups are asking for a particular stamp
        sum=sum+stamp[i][j];
    }
        summation[j]=sum;
    }

    for(int i=1;i<=N;i++){//making a graph denoting which groups cant be selected together
        if(summation[i]<=1)
        continue;
        for(int j=1;j<=M;j++){
           if(stamp[j][i]==1){

              for(int k=1;k<=M;k++)
              {
                  if(stamp[k][i]==1)
                  graph[j][k]=1;
              }
           }
        }
    }



    for(int i=1;i<=M;i++){
        int sump=0;
        possible[i]=1;
     for(int j=1;j<=M;j++)
        sump=sump+graph[i][j];
        after[i]=sump;//summing up the elements in the graphs (coloumn)
    }

    //printing the graph matrix which represents which groups cannot sit together
    /* for(int i=1;i<=M;i++){
    for(int j=1;j<=M;j++){
    printf("%d ",graph[i][j]);
    }
    printf("=%d",after[i]);
    printf("\n");
    }*/

  { //this piece finds the possible groups which can sit starting from the [row] which has lowest number of 1s because it will have least number of collisions
    int elem=M;
    while(elem>0){
    int min=999,grp=100;
    for(int i=1;i<=M;i++)
       { if(after[i]<=min)
            min=after[i],grp=i;
       }
    for(int k=1;k<=M&&grp<=M;k++)//after selecting a group from the graph matrix remove all the intersecting groups from the possible matrix
    {
        if(graph[grp][k]==1)
        possible[k]=0,elem--;
    }
        possible[grp]=1;
        after[grp]=9999;
    }
    }

   int answer=0;
   for(int i=1;i<=M;i++){
        answer=answer+possible[i];
   }
   printf("%d",answer);

   return 0;
}
//7 5 3 1 2 3 3 2 4 7 2 4 5 1 6 2 4 7
//10 7 3 1 2 4 3 1 2 4 2 1 2 1 1 2 2 4 2 5 6 4 1 5 7 8