HDU1069 Monkey and Banana

Monkey and Banana

Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u

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

题解：

给定一系列方块，可以随意翻转，问在底部面积递减的情况下，最大能堆的高度为多少，比如给定三维为abc,那么就有六种情况，abc,acb,bac,bca,cab,cba，按照底部面积从大到小排个序，然后类似LIS的做法即可

代码

#include <cstdio>#include <queue>#include <cstring>#include <iostream>#include <cstdlib>#include <algorithm>#include <vector>#include <map>#include <string>#include <set>#include <ctime>#include <cmath>#include <cctype>using namespace std;#define MAX 100000#define LL long longint cas=0,T;int dp[1001];struct mon{   int a;   int b;   int c;}monkey[1001];bool cmp(mon a1,mon b1){    if (a1.a > b1.a)        return true;    else if (a1.a == b1.a && a1.b > b1.b) return true;    else    return false;}int main(){    while (scanf("%d",&T) && T)    {      cas++;      int j = 0;      for (int i = 0;i<T;i++)      {          int tempa,tempb,tempc;          scanf("%d%d%d",&tempa,&tempb,&tempc);          monkey[j].a = tempa ;monkey[j].b = tempb; monkey[j++].c = tempc;          monkey[j].a = tempb ;monkey[j].b = tempc; monkey[j++].c = tempa;          monkey[j].a = tempc ;monkey[j].b = tempa; monkey[j++].c = tempb;          monkey[j].a = tempa ;monkey[j].b = tempc; monkey[j++].c = tempb;          monkey[j].a = tempb ;monkey[j].b = tempa; monkey[j++].c = tempc;          monkey[j].a = tempc ;monkey[j].b = tempb; monkey[j++].c = tempa;      }      sort(monkey,monkey+j,cmp);      for (int i = 0;i<T;i++)          dp[i] = monkey[i].c;      for (int i = 0;i<j;i++)          for (int k = 0;k<i;k++)          {              if (monkey[i].a < monkey[k].a && monkey[i].b < monkey[k].b && dp[i] < dp[k]+ monkey[i].c)                  dp[i] = dp[k]+monkey[i].c;          }      int maxn = 0;      for (int i = 0;i<j;i++)          if (maxn < dp[i])              maxn = dp[i];      printf("Case %d: maximum height = %d\n",cas,maxn);      memset(dp,0,sizeof(dp));      memset(monkey,0,sizeof(monkey));     }    //freopen("in","r",stdin);    //printf("time=%.3lf",(double)clock()/CLOCKS_PER_SEC);    return 0;}