Cod sursa(job #70560)

Utilizator moga_florianFlorian MOGA moga_florian Data 6 iulie 2007 13:44:34
Problema Balans Scor 0
Compilator cpp Status done
Runda Arhiva de probleme Marime 2.08 kb
#include<stdio.h>
#define Nmax 305

/*
int calc(long long X)
{
int i,j,p,q,li,lf;

//linii limita
for(p=1;p<=2*M;p++) 
  for(q=p+R-1; q<=p+M-1 && q<=2*M; q++)
     {
     //deque
     B[0]=0;
     for(j=1;j<=2*N;j++)
        B[j]= COL[q][j] - COL[p-1][j] - X*(q-p+1) + B[j-1];
        
     //intre limitele C si N
     li=lf=1;
     dq[1]=0;
     
     for(i=C;i<=2*N;i++)
        {
        if(B[i] - B[dq[li]] >= 0)
           return 1;
           
        if(i-dq[li] == N) li++;
        j= i-C+1;
        
        while(li<=lf && B[dq[lf]] > B[j]) lf--;
        dq[++lf]=j;        
        }                   
     }
     
return -1;
}
*/

int main()
{
FILE *fin=fopen("balans.in","r"),
     *fout=fopen("balans.out","w");
     
int i,j;
int COL[Nmax][Nmax],B[Nmax];
int M,N,R,C,dq[Nmax];
int A[Nmax][Nmax];

fscanf(fin,"%d%d%d%d\n",&M,&N,&R,&C);

for(i=1;i<=M;++i)
  for(j=1;j<=N;++j)
    {
    fscanf(fin,"%d",&A[i][j]);
    A[i+M][j]= A[i+M][j+N] = A[i][j+N]= A[i][j];
    }
    
   
//sume partiale
for(i=1;i<=2*M;++i)
  for(j=1;j<=2*N;++j)
     COL[i][j]= COL[i-1][j] + A[i][j];
     
      
int li=0,lf=100000000,mij,sol=0,val;
long long p,q,X;
int st,dr;

while(li<=lf)
 {
 mij=(li+lf)>>1;
 X=(long long)mij;
 
 //functia

 val=-1;
 for(p=1;p<=2*M && val<0;++p) 
  for(q=p+R-1; q<=p+M-1 && q<=2*M && val<0;++q)
     {
     //deque
     B[0]=0;
     for(j=1;j<=2*N;++j)
        B[j]= (long long)1000*(COL[q][j] - COL[p-1][j]) - X*(q-p+1) + B[j-1];
        
     //intre limitele C si N
     st=dr=1;
     dq[1]=0;
     
     for(i=C;i<=2*N;++i)
        {
        if(B[i] - B[dq[st]] >= 0)
           val=1;
           
        if(i-dq[st] == N) ++st;        
        j= i-C+1;
        
        while(st<=dr && B[dq[dr]] > B[j]) --dr;
        dq[++dr]=j;        
        }                   
     } 
  
 //
 
 if( val < 0 )
    lf=mij-1;
 else
    {
    li=mij+1;             
    sol=mij;
    }
 }
 
fprintf(fout,"%d.%03d\n",sol/1000,sol%1000);
 
fclose(fin);
fclose(fout);
return 0;         
}