hdu 1069 Monkey and Banana --)dp

Monkey and Banana

題意:

有 t 組數字,每組三個數,做為一類長方體的,長,寬,高,由於高可以是三個數字中的任意一個,所以每三個數字一般能代表3種不同類型的長方體
假設每種長方體有無限多個,問以壘積木的方式(上方的積木必須嚴格小於下方的積木),最多可以壘多高
本題可以用dp來解,先按照如果長相等寬大的放在前面,如果不等,按長由大到小排序.由於裡面存在如 (9,1) (7,8)這樣的對,而且不能保證這一段序列都是如題意所述,嚴格單調遞減的,所以不能用貪心去解答.遞推式:在編號為i,j的兩個長方體嚴格單調遞減的情況下,dp[i]=max(dp[j]+h,dp[i]).

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
0

Sample Output

Case 1: maximum height = 40
Case 2: maximum height = 21
Case 3: maximum height = 28
Case 4: maximum height = 342

code:

#include
#include
#include
#include
#include
#include
#include
using namespace std;
#define maxf(a,b) a>b?a:b
const int inf=0x3f3f3f3f;
struct node
{
    int l,w,h;
} b[1000];
bool cmp(node a, node b)
{
    if(a.l==b.l)
        return a.w>b.w;
    return a.l>b.l;
}
int main()
{
    int t,fff=1;
    while(~scanf(%d,&t)&&t)
    {
        int j=0,a[4];
        for(int i=0; i=0; i--)
        {
            for(int k=i+1; kb[k].l&&b[i].w>b[k].w)
                    dp[i]=dp[i]>dp[k]+b[i].h?dp[i]:dp[k]+b[i].h;
            }
            max1=dp[i]>max1?dp[i]:max1;
        }
        printf(Case %d: maximum height = %d
,fff++,max1);
    }
}