Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6140 Accepted Submission(s): 2618
Each area has its width and length. The area is divided into a grid of equal square units.The rent paid for each unit on which you're building stands is 3$.
Your task is to help Bob solve this problem. The whole city is divided into K areas. Each one of the areas is rectangular and has a different grid size with its own length M and width N.The existing occupied units are marked with the symbol R. The unoccupied units are marked with the symbol F.
R – reserved unit
F – free unit
In the end of each area description there is a separating line.
#include<stdio.h> #include<iostream> #include<string.h> #include<math.h> #include<algorithm> using namespace std; const int N = 1005; int n,m; char a[N][N]; int mp[N][N]; int L[N],R[N];
void input()
{
int i,j;
for(i=1; i<=n; i++)
for(j=1; j<=m; j++)
cin>>a[i][j];
for(i=1; i<=m; i++)
mp[0][i]=0;
for(i=1; i<=m; i++)
{
for(j=1; j<=n; j++)
{
if(a[j][i]=='F')
mp[j][i]=mp[j-1][i]+1;
else
mp[j][i]=0;
}
}
/*for(int i=1;i<=n;i++){ //test
for(int j=1;j<=m;j++){
printf("%d ",mp[i][j]);
}
printf("\n");
}*/
} /*void input(){ memset(mp,0,sizeof(mp)); for(int i=1;i<=n;i++){ gets(a[i]); for(int j=1;j<=m;j++){ if(i==1) { ///预处理mp if(a[i][(j-1)*2]=='R') mp[i][j]=0; if(a[i][(j-1)*2] =='F') mp[i][j]=1; }else{ if(a[i-1][(j-1)*2]=='R') { if(a[i][(j-1)*2] =='R') mp[i][j]=0; if(a[i][(j-1)*2] =='F') mp[i][j]=1; } else { if(a[i][(j-1)*2] =='R') mp[i][j]=0; if(a[i][(j-1)*2] =='F') mp[i][j]+=mp[i-1][j]+1; } } } } /*for(int i=1;i<=n;i++){ //test for(int j=1;j<=m;j++){ printf("%d ",mp[i][j]); } printf("\n"); }*/ }*/ int main() { int tcase; scanf("%d",&tcase); while(tcase--){ scanf("%d%d",&n,&m); getchar(); input(); int mx = -1; for(int k=1;k<=n;k++){ L[1]=1; R[m]=m; for(int i=2;i<=m;i++){ ///向右延伸 int t = i; if(t>1&&mp[k][i]<=mp[k][t-1]) t = L[t-1]; L[i] =t; } for(int i=m-1;i>=1;i--){ int t = i; if(t<m&&mp[k][i]<=mp[k][t+1]) t = R[t+1]; R[i] = t; } for(int i=1;i<=m;i++){ if((R[i]-L[i]+1)*mp[k][i]>mx) mx = (R[i]-L[i]+1)*mp[k][i]; } } printf("%d\n",mx*3); } }