算法竞赛入门经典 例题9-2

UVa 00437

The Tower of Babylon

有长宽高分别为x[i]、y[i]和z[i]的立方体，每种的数量有无限个，求摞起后能得到的最大高度，限制条件是上面立方体底面的长和宽严格小于下面立方体顶面的长和宽。

比较简单的一道动态规划题目，只需要将立方体依次摞上去就可以了，但是要注意的就是如何表示状态，还是需要和上一道一样进行转化，这里我将n的立方体变成了3 * n个立方体。

#include <iostream>
#include <vector>

using namespace std;

struct Cube
{
	int length, width, height;
	Cube(const int &l, const int &w, const int &h)
		:length(l), width(w), height(h) {}
};

bool placeable(int a, int b, int c, int d)
{
	return (a < c && b < d) || (a < d && b < c);
}

int calMaxHeight(const vector<Cube> &vcCube)
{
	vector<int> viHeight;
	for (auto iter = vcCube.begin(); iter != vcCube.end(); iter++)
	{
		viHeight.push_back(iter->height);
	}
	bool bUpdate = true;
	while (bUpdate) {
		bUpdate = false;
		for (size_t i = 0; i < vcCube.size(); i++)
		{
			for (size_t j = 0; j < vcCube.size(); j++)
			{
				if (i == j) continue;
				if (placeable(vcCube[j].length, vcCube[j].width,
					vcCube[i].length, vcCube[i].width)) {
					if (viHeight[j] < viHeight[i] + vcCube[j].height) {
						viHeight[j] = viHeight[i] + vcCube[j].height;
						bUpdate = true;
					}
				}
			}
		}
	}
	int maxHeight = 0;
	for (size_t i = 0; i < viHeight.size(); i++)
	{
		if (viHeight[i] > maxHeight) {
			maxHeight = viHeight[i];
		}
	}
	return maxHeight;
}

int main()
{
	int Case = 0, n;
	while (cin >> n) {
		if (n == 0) break;
		vector<Cube> vcCube;
		int x, y, z;
		for (int i = 0; i < n; i++)
		{
			cin >> x >> y >> z;
			vcCube.push_back(Cube(x, y, z));
			vcCube.push_back(Cube(x, z, y));
			vcCube.push_back(Cube(y, z, x));
		}
		cout << "Case " << ++Case << ": maximum height = ";
		cout << calMaxHeight(vcCube) << endl;
	}
	return 0;
}
/*
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
*/