【hdu 1069 Monkey and Banana（动态规划，被坑死。一遍AC）】

Monkey and Banana

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 4087 Accepted Submission(s): 2112

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

Source

University of Ulm Local Contest 1996

Recommend

JGShining

  1 // Project name : 1069 ( Monkey and Banana ) 
  2 // File name    : main.cpp
  3 // Author       : iCoding
  4 // E-mail       : honi.linux@gmail.com
  5 // Date & Time  : Wed Aug  8 14:39:00 2012
  6 
  7 
  8 #include <iostream>
  9 #include <stdio.h>
 10 #include <string>
 11 #include <cmath>
 12 #include <algorithm>
 13 using namespace std;
 14 
 15 /*************************************************************************************/
 16 /* data */
 17 #define MAXN 40
 18 
 19 class Block
 20 {
 21 public:
 22     int x;
 23     int y;
 24     int height;
 25     /*x <= y*/
 26     void manage();
 27     void setValue(int x, int y, int height);
 28     void showMap();
 29 };
 30 
 31 int n;
 32 Block iMap[MAXN*3];
 33 
 34 int iSum[MAXN*3];
 35 
 36 /*************************************************************************************/
 37 /* procedure */
 38 void Block::manage()
 39 {
 40     if (x > y)
 41     {
 42         int tmp;
 43         tmp = x;
 44         x   = y;
 45         y   = tmp;
 46     }
 47 }
 48 
 49 void Block::setValue(int x, int y, int height)
 50 {
 51     this -> x      = x;
 52     this -> y      = y;
 53     this -> height = height;
 54 }
 55 
 56 void Block::showMap()
 57 {
 58     cout << x << " " << y << " - " << height << endl;
 59 }
 60 
 61 bool cmp(Block a, Block b)
 62 {
 63     if ((a.x >= b.x && a.y >= b.y) || ((a.x * a.y) >= (b.x * b.y)))
 64     {
 65         return true;
 66     }
 67     else
 68     {
 69         return false;
 70     }
 71 }
 72 
 73 void iInit()
 74 {
 75     for (int i = 0; i < n; i++)
 76     {
 77         int a, b, c;
 78         cin >> a >> b >> c;
 79         iMap[i*3]  .setValue(a, b, c);
 80         iMap[i*3+1].setValue(b, c, a);
 81         iMap[i*3+2].setValue(c, a, b);
 82         iMap[i*3]  .manage();
 83         iMap[i*3+1].manage();
 84         iMap[i*3+2].manage();
 85     }
 86     sort(iMap, iMap + 3 * n, cmp);
 87 }
 88 
 89 void debug()
 90 {
 91     for (int i = 0; i < n * 3; i++)
 92     {
 93         iMap[i].showMap();
 94     }
 95 }
 96 
 97 bool iCanPut(Block bottom, Block up)
 98 {
 99     if (bottom.x > up.x && bottom.y > up.y)
100     {
101         return true;
102     }
103     else
104     {
105         return false;
106     }
107 }
108 
109 void iDynamicProgramming()
110 {
111     for (int i = 0; i < 3 * n; i++)
112     {
113         int iMax = 0;
114         for (int j = 0; j < i; j++)
115         {
116             if (iCanPut(iMap[j], iMap[i]) && iSum[j] > iMax)
117             {
118                 iMax = iSum[j];
119             }
120         }
121         iSum[i] = iMap[i].height + iMax;
122     }
123 }
124 
125 void iShowResult()
126 {
127     int iMax = 0;
128     for (int i = 0; i < 3 * n; i++)
129     {
130         if (iSum[i] > iMax)
131         {
132             iMax = iSum[i];
133         }
134     }
135     cout << iMax << endl;
136 }
137 
138 /*************************************************************************************/
139 /* main */
140 int main()
141 {
142     int iCaseCount = 0;
143     while (cin >> n && n)
144     {
145         iCaseCount++;
146         cout << "Case " << iCaseCount << ": maximum height = ";
147         iInit();
148         //debug();
149         iDynamicProgramming();
150         iShowResult();
151     }
152     return 0;
153 }
154 
155 // end 
156 // Code by Sublime text 2
157 // iCoding@CodeLab