Monkey and Banana（贪心+DP）

  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
  题目大意：垒积木，积木有长宽高三个特征，只有积木A的两个基础尺寸都严格小于积木B的相应基础尺寸，那么A才能放在B的上面，问你积木可以垒到的最大高度
  每次你会输入三个尺寸，而其对应的长宽高的组合有6种，我们这里先以其底座的面积排序，以面积排序，这样使得组合变成了3种，在进行DP；
  dp[i] 的含义是以i为最上面积木时的高度，初始化的话就是 dp[i] = nmb[i].z；状态转移：

dp[0] = nmb[i].z;
for(int i=1;i<n*3;i++){
	dp[i] = nmb[i].z;
	for(int j=0;j<i;j++){
		if((nmb[i].x<nmb[j].x&&nmb[i].y<nmb[j].y)||(nmb[i].y<nmb[j].x&&nmb[i].x<nmb[i].y))
			if(dp[i]<dp[j]+nmb[i].z)
				dp[i] = dp[j]+nmb[i].z;
	}
}

AC代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
const int inf = 0x3f3f3f3f;
using namespace std;
struct In{
    int x,y,z;
};
int dp[200];
bool cmp(In a,In b)
{
    return a.x*a.y>b.x*b.y;
}
int main(void)
{
    int n,mm=1;
    while(scanf("%d",&n),n){
        memset(dp,0,sizeof(dp));
        In nmb[200];
        int a,b,c,p=0;
        for(int i=1;i<=n;i++){
            scanf("%d%d%d",&a,&b,&c);
            nmb[p].x = min(a,b);
            nmb[p].y = max(a,b);
            nmb[p].z = c;

            nmb[p+1].x = min(a,c);
            nmb[p+1].y = max(a,c);
            nmb[p+1].z = b;

            nmb[p+2].x = min(c,b);
            nmb[p+2].y = max(c,b);
            nmb[p+2].z = a;

            p += 3;
        }
        sort(nmb,nmb+n*3,cmp);
        dp[0] = nmb[0].z;
        for(int i=1;i<n*3;i++){
            dp[i] = nmb[i].z;
            for(int j=0;j<i;j++){
                if((nmb[i].x<nmb[j].x&&nmb[i].y<nmb[j].y)||(nmb[i].x<nmb[j].y&&nmb[i].y<nmb[j].x)){
                    if(dp[i]<dp[j]+nmb[i].z)
                        dp[i] = dp[j]+nmb[i].z;
                }
            }
        }
        int ans = 0;
        for(int i=0;i<n*3;i++)
            if(ans<dp[i])
                ans = dp[i];
        printf("Case %d: maximum height = %d\n",mm,ans);
        mm++;
    }
    return 0;
}