In this HackerEarth Bob and cities problem solution, Bob is living in a city in which houses are arranged in NxM block.

The city is denoted by N strings having M characters such that '.' denotes house while '#' denotes forests.

Bob has to pay a certain amount LCost, RCost, UCost, DCost to move 1 step across left, right, up or down respectively.

Bob lives in a house having co-ordinates (Stx , Sty) (1-Based Indexing).

You are given Q tasks contains an integer X each. In each task, you have to find number of unique houses (including his house) can be travelled using the amount X.


HackerEarth Bob and cities problem solution


HackerEarth Bob and cities problem solution.

#include<bits/stdc++.h>
using namespace std ;
#define pb push_back
#define mp make_pair
#define infile() freopen("in12.txt","r",stdin);
#define output() freopen("out12.txt","w",stdout);
#define ll long long
#define sc(t); scanf("%d",&t);
#define scl(t); scanf("%lld",&t);
#define sc2(n,m); scanf("%d%d",&n,&m);
#define scl2(n,m); scanf("%lld%lld",&n,&m);
#define debug(); printf("tushar\n");
#define N 1001
#define mod 1000000007
#define printi(n) printf("%d",n);
#define inf ((1<<29)-1)
#define linf ((1LL<<60)-1)
const double eps = 1e-9;
set < ll > s ;
set < int > si ;
set < ll > :: iterator it ;
vector < ll > v ;
vector < int > vi ;

int n,m,q,k ;
int a[N][N] ;
char str[N][N] ;
ll dp[N][N] ;
int stx,sty ;
ll LL,RR,UU,DD ;
int valid(int x, int y)
{
if(x>=1&&x<=n&&y>=1&&y<=m&&a[x][y])
return 1 ;
return 0 ;
}


int search(ll x)
{
int low=0;
int high=v.size()-1;
int mid ;
int ans=0 ;
if(v.size() == 0)
return 0 ;
if(x<v[0])
return 0;
while(low<=high)
{
mid = (low+high)/2 ;
//printf("low = %d high = %d mid = %d x = %lld v[mid] = %lld\n",low,high,mid,x,v[mid]) ;
if(v[mid]<=x)
{
ans=mid;
low=mid+1;
}
else
high=mid-1;
}
return ans+1;
}



void func(int xx , int yy)
{
queue < pair < int , pair < int , ll > > > q ;
q.push(mp(xx,mp(yy,0))) ;
dp[xx][yy] = 0 ;
int i,j ;
while(!q.empty())
{
int x = q.front().first ;
int y = q.front().second.first ;
ll tmp = q.front().second.second ;
//printf("x = %d y = %d tmp = %lld\n",x,y,tmp) ;
q.pop() ;
int lx = x;
int ly = y-1 ;
ll cost = tmp + LL ;
if(valid(lx,ly) && cost < dp[lx][ly])
{
dp[lx][ly] = cost ;
q.push(mp(lx,mp(ly,cost))) ;
}
int rx = x;
int ry= y +1;
cost = tmp + RR ;
if(valid(rx,ry) && cost < dp[rx][ry])
{
dp[rx][ry] = cost ;
q.push(mp(rx,mp(ry,cost))) ;
}
int ux = x-1 ;
int uy = y ;
cost = tmp + UU ;
if(valid(ux,uy) && cost < dp[ux][uy])
{
dp[ux][uy] = cost ;
q.push(mp(ux,mp(uy,cost))) ;
}
int dx = x+1 ;
int dy = y ;
cost = tmp + DD ;
if(valid(dx,dy) && cost < dp[dx][dy])
{
dp[dx][dy] = cost ;
q.push(mp(dx,mp(dy,cost))) ;
}
}

for(i=1;i<=n;i++)
{
for(j=1;j<=m;j++)
{
if(dp[i][j]!=linf)
v.pb(dp[i][j]) ;
}
}
sort(v.begin(),v.end());
}
int main()
{
int i , j , t ;

//infile() ;
//output() ;

sc2(n,m) ;

for(i=0;i<n;i++)
scanf("%s",str[i]) ;

scanf("%lld %lld %lld %lld",&LL,&RR,&UU,&DD );
//printf("LL = %lld RR = %lld UU = %lld DD = %lld\n",LL,RR,UU,DD) ;
for(i = 1 ; i <= n ; i++ )
{
for(j=1;j<=m;j++)
{
if(str[i-1][j-1] == '.')
a[i][j] = 1 ;
else
a[i][j] = 0 ;
}
}

for(i=1;i<=n;i++)
for(j=1;j<=m;j++)
dp[i][j]=linf ;

sc2(stx,sty) ;
func(stx,sty) ;

ll petrol ;
int ans = 0 ;
for(i=1;i<=n;i++)
{
for(j=1;j<=m;j++)
{
if(a[i][j] == 1)
{
if(dp[i][j] <= petrol)
ans++ ;
}
}
}


//printf("sz = %d printf"("jhdsb")"\n",v.size()) ;

sc(q) ;
while(q--)
{
ll pet ;
scl(pet) ;
printf("%d\n",search(pet)) ;
}
//printf("%d\n",ans) ;
return 0 ;
}