hdu1069java

給你n個方塊，其中每個方塊具有它的長寬高（方塊可以任意旋轉放置），方塊數量不限。現在你要堆一個高塔，上面方塊的長和寬必須嚴格小於下面方塊的長和寬。問你能堆起來的最大高度。
先將方塊以長和寬按從小到大排序，然後從小到大以此為底，求出最大高度。dp[i] = max(dp[j])+i.height (j.x<i.x&&j.y<i.y)，最後的結果為max(dp)。

import java.util.Arrays;
import java.util.Scanner;

class Block implements Comparable<Block>{
    int x, y, z;
    Block(int _x, int _y, int _z) {
        x = _x; y = _y; z = _z;
    }
    @Override
    public int compareTo(Block arg0) {
        if(x != arg0.x)
            return x - arg0.x;
        return y - arg0.y;
    }    
}

public class hdu1069 {

    public static void main(String[] args) {
        // TODO 自動生成的方法存根
        Scanner sc = new Scanner(System.in);
        int cas = 1;
        while (sc.hasNext()) {
            int n = sc.nextInt();
            if (n==0) {
                break;
            }
            Block[] bk = new Block[6*n];
            for (int i = 0; i < n; i++) {
                int x = sc.nextInt();
                int y = sc.nextInt();
                int z = sc.nextInt();
                bk[6*i] = new Block(x, y, z);
                bk[6*i+1] = new Block(x, z, y);
                bk[6*i+2] = new Block(y, x, z);
                bk[6*i+3] = new Block(y, z, x);
                bk[6*i+4] = new Block(z, x, y);
                bk[6*i+5] = new Block(z, y, x);
            }
            Arrays.sort(bk);
            int[] dp = new int[6*n];
             //設以誰為底
            int ans = 0;
            for (int i = 0; i < bk.length; i++) {
                int max = 0;
                for (int j = 0; j < i; j++) {
                    if ((bk[j].x >= bk[i].x)||(bk[j].y >= bk[i].y)) {
                        continue;
                    }
                    max = Math.max(max, dp[j]);
                }
                dp[i] = max+bk[i].z;
                ans = Math.max(ans, dp[i]);
            }
            System.out.println("Case "+cas+": maximum height = "+ans);
            cas++;
            
        }
        sc.close();
    }
}