Philosophers Stone
Given below code is for BYTESM2 spoj or Philosophers Stone spoj
Hint:- Simple DP
#include <iostream>
#include <cstdlib>
#include <algorithm>
using namespace std;
int main()
{
int t;
cin>>t;
while(t--)
{
int h , m;
cin>>h>>m;
int arr[h+9][m+9];
for(int i = 0 ; i < h ; i++)
for(int j = 0 ; j < m ; j++)
cin>>arr[i][j];
int ma = -1;
for(int i = 0 ; i < m ; i++)
ma = max(ma , arr[0][i]);
for(int i = 1 ; i < h ; i++)
{
ma = -1;
for(int j = 0 ; j < m ; j++)
{
if(j>0 && j<m-1)
arr[i][j] = max(arr[i-1][j] + arr[i][j] , max(arr[i-1][j-1]+arr[i][j] , arr[i-1][j+1]+arr[i][j]));
else if(j>0)
arr[i][j] = max(arr[i-1][j]+arr[i][j] ,arr[i-1][j-1]+arr[i][j]);
else if(j<m-1)
arr[i][j] = max(arr[i-1][j]+arr[i][j],arr[i-1][j+1]+arr[i][j]);
ma = max(arr[i][j] , ma);
}
}
cout<<ma<<endl;
}
return 0;
}
thanks for the code...
ReplyDelete#include
ReplyDeleteusing namespace std;
int findmax(int a,int b,int c)
{
int max=a;
if(max>d[i][j];
for(j=1;j<=w;j++)
dp[1][j]=d[1][j];
//storing inf value
for(i=1;i<h;i++)
dp[i][0]=-1;
//recurrence relation
for(i=2;i<=h;i++)
for(j=1;j<=w;j++)
dp[i][j]=d[i][j]+findmax(dp[i-1][j],dp[i-1][j+1],dp[i-1][j-1]);
//answer
int ans=dp[h][1];
for(j=2;j<=w;j++)
if(ans<dp[h][j])
ans=dp[h][j];
printf("%d\n",ans);
}
}
Why this answer is wrong
#include
ReplyDeleteusing namespace std;
int findmax(int a,int b,int c)
{
int max=a;
if(max>d[i][j];
for(j=1;j<=w;j++)
dp[1][j]=d[1][j];
//storing inf value
for(i=1;i<h;i++)
dp[i][0]=-1;
//recurrence relation
for(i=2;i<=h;i++)
for(j=1;j<=w;j++)
dp[i][j]=d[i][j]+findmax(dp[i-1][j],dp[i-1][j+1],dp[i-1][j-1]);
//answer
int ans=dp[h][1];
for(j=2;j<=w;j++)
if(ans<dp[h][j])
ans=dp[h][j];
printf("%d\n",ans);
}
}