题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1069
Monkey and Banana
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5272 Accepted Submission(s): 2702
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.
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.
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
解题思路：这题是一个最长递减序列的变式，题目要求是把石块堆起来，下面的那块要比上面的要大（即长和宽都大于上面的那块），那么我们只需排好序就可以操作了，求最长的递减序列和！
解题代码：
View Code G++
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <cmath> 5 #include <algorithm> 6 using namespace std; 7 typedef __int64 LL; 8 struct node 9 { 10 LL x, y, z; 11 bool operator < (const node a) const 12 { 13 if (x == a.x) 14 return y > a.y; 15 return x > a.x; 16 } 17 }sto[300]; 18 LL dp[300]; 19 20 bool cmp1(node &a, node&b) 21 { 22 if(a.x < b.x && a.y < b.y) 23 return true; 24 return false; 25 } 26 27 int main () 28 { 29 int n, i, j; 30 int Case = 1; 31 while (scanf ("%d", &n) == 1, n) 32 { 33 int cun = 1; 34 sto[0] = (node){1e9, 1e9, 0}; 35 memset (dp, 0, sizeof (dp)); 36 for (i = 0; i < n; i ++) 37 { 38 LL x, y, z; 39 scanf ("%I64d%I64d%I64d", &x, &y, &z); 40 sto[cun ++] = (node) {x, y, z}; 41 sto[cun ++] = (node) {x, z, y}; 42 sto[cun ++] = (node) {y, x, z}; 43 sto[cun ++] = (node) {y, z, x}; 44 sto[cun ++] = (node) {z, x, y}; 45 sto[cun ++] = (node) {z, y, x}; 46 } 47 sort (sto+1, sto+cun); 48 LL ans = 0; 49 for (i = 1; i < cun; ans = max (ans, dp[i]), i ++) 50 { 51 for (j = 0; j < i; j ++) 52 { 53 if (cmp1(sto[i], sto[j]) && dp[j] + sto[i].z > dp[i]) 54 dp[i] = dp[j] + sto[i].z; 55 } 56 } 57 printf ("Case %d: maximum height = %I64d\n", Case++, ans); 58 } 59 return 0; 60 }