problem with implementing this question of codechef

[needbro]

please pm me

[ghostrideriit]

@MonkeyD

Can u discuss the algo to solve TWO Flowers for 20 points....

[ak2614]

@needbro please pm your code i will try to rectify it....

[noobiee123]

I have passed subtask 1 but not subtask 2.. any possible errors? TIA

[closed account 3UohqMoL]

@noobiee how can we point out your errors without your algorithm.

[keve]

@ wasp
have you solved two flowers problem
if yes please give me the idea of how to approach for this problem

[needbro]

thanks @mythio and @ak2614 for providing the needed help:)

[keve]

hello guys
can anyone help me with the two flowers partial solution
please help me.

[hyfun]

@texasgirl1234
refer connected cell problem hackerrank

[ghostrideriit]

@all

In two flower problem u can collect 1 and max. 2 flower from a cell and these should be in continous if adjacent to it are more than 2 flower than break from there so like this we have to counter the total no. Of cells not flowers from which we can collect flowers either 1 or 2.........

[yobro]

please someone help me rectify my output
giving wrong answer for original constraints

long double t,t1,t2;

if(X==0){
t=(-1*Z)/Y;
cout<<setprecision(12)<<t<<endl;
}
else{
t1=((-1*Y)+sqrt(Y*Y-(4*X*Z)))/(2*X);
t2=((-1*Y)-sqrt(Y*Y-(4*X*Z)))/(2*X);

if(t1>=0 && t2<0)cout<<setprecision(12)<<t1<<endl;
else if(t1<0 && t2>=0)cout<<setprecision(12)<<t2<<endl;
else if(t1>=0 && t2>=0)cout<<setprecision(12)<<min(t1,t2)<<endl;
}

[iitkgpian]

#include <stdio.h>
#include<math.h>
int n,m,ct;
int a[100][100];
void check(int i,int j)
{
if(a[i][j]==0)
return;
a[i][j]=5;
ct++;
if(i+1<n&&a[i+1][j]==1 || a[i+1][j]==2){
check(i+1,j);

}
if(i-1>=0&&a[i-1][j]==1 || a[i-1][j]==2){
check(i-1,j);

}
if(j+1<m&&a[i][j+1]==1 || a[i][j+1]==2){
check(i,j+1);

}
if(j-1>=0&&a[i][j-1]==1 || a[i][j-1]==2){
check(i,j-1);

}

}
int main() {
while(scanf("%d%d",&n,&m)!=EOF)
{
int tt;
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{tt=scanf("%d",&a[i][j]);}//printf("%d ",a[i][j]);}
//puts("");
}
//continue;
int ans=0;
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
if(a[i][j]==1 || a[i][j]==2)
{
ct=0;
check(i,j);
if(ct>ans) ans=ct;
}
}
}
printf("%d\n",ans);
}
return 0;
}
Just passing 2 test case only.
@lastchance @Wase please see where I am getting wrong in 2 flowers problem
Any help would be appreciated

[iitkgpian]

@yobro
Which question is it??
If it is vision one that this may be wrong if you have calculated the derived constants wrongly.
Try to post full code for review.

[closed account 1TD80pDG]

anyone can help for this problem.
Two Flowers Problem Code: TWOFL
its some kind of similar to this problem at hackerrank but problem is that in hackrank we consider diagonal (where cell 1 only) but in codechef we need only consider side adjecent of 1 or 2.
we can convert 1 and 2 to 1 and rest to Zero to make same as hackrank but how can i remove diagonal condition in problems in given code .

#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
using namespace std;

int m[100][100],t=2;

void region(int r,int c){
    
   m[r][c]=t;
    for(int i=r-1;i<=r+1;i++){
        for(int j=c-1;j<=c+1;j++){
            if( m[i][j]==1){
                m[i][j]=t;
                region(i,j);
                
            }
        }
    }
    
    return;
}
    
int main() { 
    int r,c;
    cin>>r>>c;
    
    for(int i=0;i<r;i++)
        for(int j=0;j<c;j++)
        {  cin>>m[i][j];
        
           if(m[i][j]==1 ||m[i][j]==2)
               m[i][j]=1;
         else
             m[i][j]=0;
         
        }
    
    
        
        for(int i=0;i<r;i++){
        for(int j=0;j<c;j++){
        if(m[i][j]==1){
        region(i,j);
           t++;
         }         
        }            
    }    
    int ar[t+1]={0};
    for(int i=0;i<r;i++){
        for(int j=0;j<c;j++)
        ar[m[i][j]]+=1;
    }
    int max=0;
    for(int i=1;i<=t;i++)
        max=fmax(max,ar[i]);
    
    cout<<max;
        
    return 0;
}

https://www.hackerrank.com/challenges/connected-cell-in-a-grid/problem

[closed account 1TD80pDG]

anyone wants the first four problems full 100% AC answer. You need only give me any partial or fully solved answer form the last four problem (after the vision problem) . interested students PM me.

[keve]

can anyone tell that
what are changes have to made in the solution of @iitkgpian to get all AC

uptil now it is giving -
1 0 WA (0.000000)
1 1 AC (0.000000)
1 2 WA (0.000000)
1 3 WA (0.000000)
1 4 AC (0.000000)
Subtask Score: 0.00% Result - WA
2 5 WA
(0.140000)
2 6 AC
(0.140000)
2 7 WA
(0.220000)
2 8 WA
(0.220000)
2 9 AC
(0.110000)
Subtask Score: 0.00% Result - WA
3 10 WA
(0.550000)
3 11 AC
(0.620000)
3 12 WA
(0.520000)
3 13 WA
(0.520000)
3 14 AC
(0.500000)
Subtask Score: 0.00% Result - WA
Total Score = 0.00%

[closed account 1TD80pDG]

anyone convert this code in such a way that it read two intiger in single line
and the value reading for matrixes if(a[i][j]==1 ||a[i][j]==2) then a[i][j]=1; else a[i][j]=0;
it gives me write answer if take input this way
4
5
1 1 2 3 1
3 1 2 5 2
5 2 1 5 6
1 3 1 2 1

but i want to take input this way
4 5
1 1 2 3 1
3 1 2 5 2
5 2 1 5 6
1 3 1 2 1

and then i want to convert 1 to 1 and 2to 1 and rest to Zero(0).

this way
4 5
1 1 1 0 1
0 1 1 0 1
0 1 1 0 0
1 0 1 1 1


anyone know python can convert this. then it gives correct answer .

#!/bin/python3

import math
import os
import random
import re
import sys

# Complete the connectedCell function below.
#def connectedCell(matrix):
def connectedCell(matrix):
    # Complete this function
    result_list = []
    n,m = len(matrix),len(matrix[0])
    for i in range(0,n):
        for j in range(0,m):
            if matrix[i][j] == 1:
                result_list.append(dfs(matrix,i,j))
    return max(result_list)
                
def dfs(matrix,i,j):
    n,m = len(matrix),len(matrix[0])
    if i < 0 or j < 0 or i >= n or j >= m:
        return 0
    if matrix[i][j] == 0:
        return 0
    if matrix[i][j] == 1:
        matrix[i][j] = 0
        return 1+dfs(matrix,i-1,j)+dfs(matrix,i+1,j)+dfs(matrix,i,j-1)+dfs(matrix,i,j+1) 

if __name__ == '__main__':
    fptr = open(os.environ['OUTPUT_PATH'], 'w')

    
   
    n = int(input())

    m = int(input())

  # n,m = map(int,sys.stdin.readline().split())

    matrix = []

    for _ in range(n):
        matrix.append(list(map(int, input().rstrip().split())))

    result = connectedCell(matrix)

    fptr.write(str(result) + '\n')

    fptr.close()

[yobro]

@keve
please check ur pm

[closed account 3UohqMoL]

@yobro pm me too

[yobro]

@wasp done
please help