C

（）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

题意：给出很多种箱子，每种箱子有三个参数 a,b,c（长、宽、高），每种箱子有五无限个。当且仅当 上面的箱子长严格小于下面的箱子，上面箱子的宽严格小于下面的箱子，我们可以将小箱子放置上去。求：我们最高可以叠到多高。

PS：每一个箱子，根据六个面的摆放姿势，可以看做六种箱子（不严格定义长宽），也可以看做三种箱子（严格定义长宽）。

#include<iostream>#include<cstring>#include<algorithm>using namespace std;#define ll long long int #define INF 0x3f3f3f3f#define Irish_Moonshine mainconst int maxn = 205;int dp[maxn];struct node {    int x, y, z;    bool operator <(const node t)const//对于底部长进行一波排序    {        return x > t.x;    }}blocks[maxn];int cnt;void add(int x, int y, int z){    blocks[cnt].x = x;    blocks[cnt].y = y;    blocks[cnt].z = z;    cnt++;}int Irish_Moonshine(){    int n, cs = 0;    while (~scanf("%d", &n) && n)    {        cnt = 0;        int a, b, c;        int ans = 0;        for (int i = 1; i <= n; i++)        {            scanf("%d%d%d", &a, &b, &c);            add(a, b, c);            add(a, c, b);            add(b, a, c);            add(b, c, a);            add(c, a, b);            add(c, b, a);        }        sort(blocks, blocks + cnt);        memset(dp, 0, sizeof(dp));        for (int i = 0; i < cnt; i++)        {            dp[i] = blocks[i].z;//lowest            for (int j = 0; j < i; j++)//将高度最高的放置在i下面            {                if (blocks[j].x > blocks[i].x&&blocks[j].y > blocks[i].y)                    dp[i] = max(dp[i], dp[j] + blocks[i].z);                //从j的位置上再放置一个i~~~            }            ans = max(dp[i], ans);        }        printf("Case %d: maximum height = %d\n", ++cs, ans);    }    return 0;}