UVa 437 The Tower of Babylon（简单DP）

题意：

给你n种石头，长x，宽y，高z，每种石头数目无限，一块石头能放到另一块上的条件是：长和宽严格小于下面的石头。问叠起来的最大高度。

思路：

dp[i]表示选择第i个能达到的最大高度。注意每个石头可以有6种组合，都要输入进去。类似于LIS。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>

using namespace std;

int dp[256];

struct Node {
    int x, y, z;

    friend bool operator < (const Node &a, const Node &b) {
        if (a.x == b.x)
            return a.y > b.y;
        return a.x > b.x;
    }
} node[256];

bool judge(const Node &a, const Node &b)
{
    if (a.x < b.x && a.y < b.y)
        return true;
    return false;
}

int main()
{
    int n, count = 0;
    while (scanf("%d", &n) && n)
    {
        for (int i = 0; i < n; ++i)
        {
            int a, b, c;
            scanf("%d %d %d", &a, &b, &c);
            node[i*6].x = a, node[i*6].y = b, node[i*6].z = c;
            node[i*6+1].x = a, node[i*6+1].y = c, node[i*6+1].z = b;
            node[i*6+2].x = b, node[i*6+2].y = a, node[i*6+2].z = c;
            node[i*6+3].x = b, node[i*6+3].y = c, node[i*6+3].z = a;
            node[i*6+4].x = c, node[i*6+4].y = a, node[i*6+4].z = b;
            node[i*6+5].x = c, node[i*6+5].y = b, node[i*6+5].z = a;
        }
        sort(node, node + n * 6);

        int ans = 0;
        memset(dp, 0, sizeof(dp));
        dp[0] = node[0].z;
        for (int i = 1; i < n * 6; ++i)
        {
            dp[i] = node[i].z;
            for (int j = 0; j < i; ++j)
                if (judge(node[i], node[j]) && dp[i] < dp[j] + node[i].z)
                    dp[i] = dp[j] + node[i].z;

            if (dp[i] > ans)
                ans = dp[i];
        }
        ++count;
        printf("Case %d: maximum height = %d\n", count, ans);
    }
    return 0;
}