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;

}