problem with implementing this question

@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 (6)

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 (9)

#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

Last edited on

iitkgpian (9)

@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.

Last edited on

closed account 1TD80pDG (101)

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

Last edited on

closed account 1TD80pDG (101)

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.

Last edited on

keve (21)

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%

Last edited on

closed account 1TD80pDG (101)

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()

Last edited on

yobro (6)

@keve
please check ur pm

closed account 3UohqMoL (121)

@yobro pm me too

yobro (6)

@wasp done
please help

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problem with implementing this question of codechef