UVa - 437 - The Tower of Babylon

Perhaps you have heard of the legend of the Tower of Babylon. Nowadays many details of this tale have been forgotten. So now, in line with the educational nature of this contest, we will tell you the whole story:

The babylonians had n types of blocks, and an unlimited supply of blocks of each type. Each type-i block was a rectangular solid with linear dimensions  . 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 wanted to construct the tallest tower possible by stacking blocks. The problem was 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. 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 babylonians can build with a given set of blocks.

Input and Output

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  ,  and  .

Input is terminated by a value of zero (0) for n.

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

动态规划,注意如果用d(a, b)表示状态值(顶面尺寸为a*b),不好,因为a和b太大会导致数组开不下,而且浪费了太多的内存,很多数组值都没有用到。所以换一个状态。用(idx, k)表示状态,idx为顶面立方体的序号,k是高的序号(把每个立方体的3个纬度从小到大排序,编号为0~2)。时间复杂度O(n^2)

AC代码:

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include  
#include  
#include 
#include 

using namespace std;

const int maxn = 35;

int n;
int blocks[maxn][3];
int d[maxn][3];
// 状态数组,d[idx][k]当顶面为立方体idx,其中第k条边(排序后)为高

void init()
{
	memset(d, 0, sizeof(d));
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < 3; j++) {
			cin >> blocks[i][j];
		}
		sort(blocks[i], blocks[i] + 3);
	}
}

void getDimensions(int* v, int b, int dim)
{
	int idx = 0;
	for (int i = 0; i < 3; i++) {
		if (i != dim) {
			v[idx++] = blocks[b][i];
		}
	}
}

int dp(int i, int j)
{
	int& ans = d[i][j];
	if (ans > 0) {
		return ans;
	}
	ans = 0;
	int v[2], v2[2];
	getDimensions(v, i, j);
	for (int a = 0; a < n; a++) {
		for (int b = 0; b < 3; b++) {
			getDimensions(v2, a, b);
			if (v2[0] < v[0] && v2[1] < v[1]) {
				ans = max(ans, dp(a, b));
			}
		}
	}
	ans += blocks[i][j];
	return ans;
}

int main()
{
	ios::sync_with_stdio(false);
	int kase = 0;
	while (cin >> n && n) {
		init();
		int ans = 0;
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < 3; j++) {
				ans = max(ans, dp(i, j));
			}
		}
		cout << "Case " << ++kase << ": maximum height = " << ans << endl;
	}

	return 0;
}

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