HDU 1069：Monkey and Banana（DP）
Monkey and Banana
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 19928 Accepted Submission(s): 10609
Problem Description
A group of researchers are designing an experiment to 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.
Input
The input file will contain one or more test cases. The 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.
Output
For each test case, print one line containing the case number (they are numbered sequentially starting from 1) and the height of the tallest possible tower in the format "Case case: maximum height = height".
Sample Input
1
10 20 30
2
6 8 10
5 5 5
7
1 1 1
2 2 2
3 3 3
4 4 4
5 5 5
6 6 6
7 7 7
5
31 41 59
26 53 58
97 93 23
84 62 64
33 83 27
0Sample Output
Case 1: maximum height = 40
Case 2: maximum height = 21
Case 3: maximum height = 28
Case 4: maximum height = 342题意
有n种砖块，每种砖块的长宽高已知，每种砖块个数无限，选出一些砖块建塔，要求在下面砖块的长和宽大于上面砖块的长宽，在满足这个要求的情况下，塔的最大高度
思路
每次输入一种砖块的长宽高，把这种砖的摆放情况情况枚举出来，存到结构体中，然后dp就行了
AC代码
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <math.h>
#include <limits.h>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <set>
#include <string>
#define ll long long
#define ms(a) memset(a,0,sizeof(a))
#define pi acos(-1.0)
#define INF 0x3f3f3f3f
const double E=exp(1);
const int maxn=1e3+10;
using namespace std;
struct wzy
{
int a,b,c;//长宽高
int dp;
}p[maxn];
bool cmp(wzy u,wzy v)
{
if(u.a==v.a)
return u.b<v.b;
return u.a<v.a;
}
int main(int argc, char const *argv[])
{
ios::sync_with_stdio(false);
int n;
int t=0;
int x,y,z;
while(cin>>n&&n)
{
ms(p);
int k=0;
for(int i=0;i<n;i++)
{
cin>>x>>y>>z;
if(x==y)
{
if(x==z)
{
p[k].a=x;p[k].b=y;p[k].c=z;p[k].dp=p[k].c;k++;
}
else
{
p[k].a=x;p[k].b=y;p[k].c=z;p[k].dp=p[k].c;k++;
p[k].a=x;p[k].b=z;p[k].c=y;p[k].dp=p[k].c;k++;
p[k].a=z;p[k].b=x;p[k].c=y;p[k].dp=p[k].c;k++;
}
}
else
{
if(y==z)
{
p[k].a=x;p[k].b=y;p[k].c=z;p[k].dp=p[k].c;k++;
p[k].a=y;p[k].b=z;p[k].c=x;p[k].dp=p[k].c;k++;
p[k].a=z;p[k].b=x;p[k].c=y;p[k].dp=p[k].c;k++;
}
else if(x==z)
{
p[k].a=x;p[k].b=y;p[k].c=z;p[k].dp=p[k].c;k++;
p[k].a=x;p[k].b=z;p[k].c=y;p[k].dp=p[k].c;k++;
p[k].a=y;p[k].b=x;p[k].c=z;p[k].dp=p[k].c;k++;
}
else
{
p[k].a=x;p[k].b=y;p[k].c=z;p[k].dp=p[k].c;k++;
p[k].a=x;p[k].b=z;p[k].c=y;p[k].dp=p[k].c;k++;
p[k].a=y;p[k].b=x;p[k].c=z;p[k].dp=p[k].c;k++;
p[k].a=y;p[k].b=z;p[k].c=x;p[k].dp=p[k].c;k++;
p[k].a=z;p[k].b=y;p[k].c=x;p[k].dp=p[k].c;k++;
p[k].a=z;p[k].b=x;p[k].c=y;p[k].dp=p[k].c;k++;
}
}
}
sort(p,p+k,cmp);
int ans=0;
for(int i=1;i<k;i++)
{
for(int j=0;j<i;j++)
{
if(p[i].a>p[j].a&&p[i].b>p[j].b)
p[i].dp=max(p[i].dp,p[j].dp+p[i].c);
}
ans=max(ans,p[i].dp);
}
cout<<"Case "<<++t<<": maximum height = ";
cout<<ans<<endl;
}
return 0;
}
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