POJ-1190 Birthday Cake
Description:
July 17th is Mr. W's birthday. For this purpose, ACM-THU will make an M-layer birthday cake with a volume of Nπ, each layer is a cylinder.
Suppose the i-th layer of cake from bottom to top (1 <= i <= M) is a cylinder with radius Ri and height Hi. When i <M, it is required that Ri> Ri+1 and Hi> Hi+1.
Due to the need to put cream on the cake, in order to save money as much as possible, we hope that the area Q of the outer surface of the cake (except the bottom surface of the bottom layer) is the smallest.
Let Q = Sπ
, please program to find out the cake making plan (appropriate values of Ri and Hi) for the given N and M, so that S is the smallest.
(Except Q, all the above data are positive integers)
Input
has two lines, the first line is N (N <= 10000), which means the volume of the cake to be made is Nπ; the second line is M (M <= 20), Indicates that the number of layers of the cake is M.
Output
is only one line and is a positive integer S (if there is no solution, then S = 0).
Sample Input
100
2
Sample Output
68
Hint
cylindrical formula
Volume V = πR2H
side area A'= 2πRH
bottom area A = πR2
Sponsor
question analysis: Direct search is obviously unrealistic, we need pruning, then we can think from the following aspects:
1：当前面积比最小面积大，可以剪枝
2：当前面积加上剩下最小面积大于最小面积，可以剪枝
3：若剩余的体积比做最大的蛋糕的体积还要大，那也可以剪枝了，
code show as below:
#include <stdio.h>
void DFS(int e,int V,int C,int R,int H);
int n,m,k=0;
int main()
{
scanf("%d%d",&n,&m);
DFS(m,n,0,100,1000);
printf("%d",k);
return 0;
}
void DFS( int e, int V, int C, int R, int H)
{
int r,h,v,c;
if(V<0)
{
return;
}
if(!e)
{
if(!V&& (C<k||!k))
{
k=C;
}
return;
}
if(k&&C>k)
{
return;
}
if(e*(R-1)*(R-1)*(H-1)<V &&e!=m)
{
return;
}
for(r=R-1;r>=e;r--)
{
for(h=H-1;h>=e;h--)
{
c=2*r*h;
v=r*r*h;
if((C+((2*V)/r))>k&&k)
{
continue;
}
if(e==m)
{
c+=r*r;
}
DFS(e-1,V-v,C+c,r,h);
}
}
}