DP HDU-1069

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

  这个题目一开始以为是数据错误了，结果最后发现是没有看见每个方块可以用无限个，而且长宽高可以互换。所以以后一定要认真读题。这题也是一道基础的DP。

最重要的环节是每次输入一个长方体快的数据，就要在原有的地方加上它所可能的六种组合。

#include <iostream>#include <stdio.h>#include <cstring>#include <algorithm>using namespace std;#define Max_N 305struct rectangle{  int l, w, h;}r[Max_N];int dp[Max_N];int N;int a, b,c;int m ;bool cmp(rectangle a, rectangle b){  if (a.w < b.w) return true;  else if (a.w == b.w && a.l < b.l) return true;  // 这里是定义一个顺序，对所有的方块进行排序。  return false;}bool cmp1(rectangle a, rectangle b){  if (a.l > b.l && a.w > b.w) return true;  return false;}int main (){  int w = 0;  while (true) {    scanf("%d", &N);    if (N == 0) break;    w ++;    m = 0;    for (int i = 0; i < N; i++)    {      scanf("%d%d%d", &a, &b, &c);                 // 这里就是做了数据处理，此步可以做优化。      r[m + 0].l = a; r[m + 0].w = b; r[m + 0].h = c;      r[m + 1].l = a; r[m + 1].w = c; r[m + 1].h = b;      r[m + 2].l = b; r[m + 2].w = a; r[m + 2].h = c;      r[m + 3].l = b; r[m + 3].w = c; r[m + 3].h = a;      r[m + 4].l = c; r[m + 4].w = a; r[m + 4].h = b;      r[m + 5].l = c; r[m + 5].w = b; r[m + 5].h = a;      m += 6;  }    sort (r, r + m, cmp);    for (int i = 0; i < m; i++)   // 初始化dp[i]的数据。      dp[i] = r[i].h;    for (int i = 0; i < m; i++)      for (int j = 0; j < i; j++) {        if (r[i].l > r[j].l && r[i].w > r[j].w) {          dp[i] = max(dp[i], dp[j] + r[i].h);        }      }    int max1 = 0;    for (int i = 0; i < m ; i++) {      if (max1 < dp[i]) max1 = dp[i];  }    printf("Case %d: maximum height = %d\n", w,max1);  }  return 0;}