Monkey and Banana
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6866 Accepted Submission(s):
3516
test the IQ of a monkey. They will hang a banana at the roof of a building, and
at the mean time, provide the monkey with some blocks. If the monkey is clever
enough, it shall be able to reach the banana by placing one block on the top
another to build a tower and climb up to get its favorite food.
The
researchers have n types of blocks, and an unlimited supply of blocks of each
type. Each type-i block was a rectangular solid with linear dimensions (xi, yi,
zi). A block could be reoriented so that any two of its three dimensions
determined the dimensions of the base and the other dimension was the height.
They want to make sure that the tallest tower possible by stacking
blocks can reach the roof. The problem is that, in building a tower, one block
could only be placed on top of another block as long as the two base dimensions
of the upper block were both strictly smaller than the corresponding base
dimensions of the lower block because there has to be some space for the monkey
to step on. This meant, for example, that blocks oriented to have equal-sized
bases couldn't be stacked.
Your job is to write a program that
determines the height of the tallest tower the monkey can build with a given set
of blocks.
first line of each test case contains an integer n,
representing the number
of different blocks in the following data set. The maximum value for n is
30.
Each of the next n lines contains three integers representing the values
xi, yi and zi.
Input is terminated by a value of zero (0) for n.
number (they are numbered sequentially starting from 1) and the height of the
tallest possible tower in the format "Case case: maximum height =
height".
#include <iostream>
#include <algorithm>
using namespace std; typedef struct rectangular{
int x,y,z;
}R;
R r[];
int k;
bool cmp(R a,R b)
{
if(a.x == b.x)
return a.y<b.y;
return a.x<b.x;
}
int dp()
{
int i, j;
int dp[];
sort(r,r+k,cmp);
int maxheight = ;
for(i=; i<k; i++)
{
dp[i] = r[i].z;
for(j=;j<i; j++)
if(r[i].x>r[j].x && r[i].y>r[j].y)
if(dp[j]+r[i].z > dp[i])
dp[i] = dp[j]+r[i].z;
if(maxheight < dp[i])
maxheight = dp[i];
}
return maxheight;
}
int main()
{
int n, i, cas=;
int x, y, z;
while(cin>>n && n)
{
k = ;
for(i=; i<n; i++)
{
cin>>x>>y>>z;
r[k].x = max(x,y);
r[k].y = min(x,y);
r[k++].z = z;
r[k].x = max(x,z);
r[k].y = min(x,z);
r[k++].z = y;
r[k].x = max(y,z);
r[k].y = min(y,z);
r[k++].z = x;
}
cout<<"Case "<<cas++<<": maximum height = "<<dp()<<endl;
}
return ;
}