Monkey and Banana（DP）

Monkey and Banana

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7416    Accepted Submission(s): 3813

Problem Description

A group of researchers are designing an experiment to test the IQ of a monkey. They will hang a banana at the roof of a building, and at the mean time, provide the monkey with some blocks. If the monkey is clever enough, it shall be able to reach the banana by placing one block on the top another to build a tower and climb up to get its favorite food.

The researchers have n types of blocks, and an unlimited supply of blocks of each type. Each type-i block was a rectangular solid with linear dimensions (xi, yi, zi). 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 want to make sure that the tallest tower possible by stacking blocks can reach the roof. The problem is 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 because there has to be some space for the monkey to step on. 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 monkey can build with a given set of blocks.

 

 

Input

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 xi, yi and zi.
Input is terminated by a value of zero (0) for n.

 

 

Output

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

 

     题意：

     给出 n （n <= 30），代表给出 n 种长方体，给出每个长方体的三个长度，可以任意摆放任意使用多个，问叠放起来的最高高度是多少，输出最高的高度。底部的两边必须严格小于下面的底部才能往上叠放。

 

     思路：

     DP。类似于矩形嵌套的变形。找最长上升子序列。将一个长方体拆开六种来判断即可，dp[ i ] 表示到第 i 个长方体的最高高度是多少。

 

     AC：

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

using namespace std;

typedef struct { int l, w, h; } node;

node num[200];
int ans, dp[200];

bool cmp (node a, node b) {
    if (a.l != b.l) return a.l > b.l;
    if (a.w != b.w) return a.w > b.w;
    return a.h > b.h;
}

int main() {

    int n, t = 0;
    while (~scanf("%d", &n) && n) {
        ans = 0;
        memset(dp, 0, sizeof(dp));

        while (n--) {
            int a, b, c;
            scanf("%d%d%d", &a, &b, &c);
            num[ans].l = a, num[ans].w = b, num[ans++].h = c;
            num[ans].l = a, num[ans].w = c, num[ans++].h = b;
            num[ans].l = b, num[ans].w = a, num[ans++].h = c;
            num[ans].l = b, num[ans].w = c, num[ans++].h = a;
            num[ans].l = c, num[ans].w = a, num[ans++].h = b;
            num[ans].l = c, num[ans].w = b, num[ans++].h = a;
        }

        sort(num, num + ans, cmp);

        int max_high = 0;
        dp[0] = num[0].h;
        for (int i = 1; i < ans; ++i) {
            int max_temp = 0;
            for (int j = 0; j < i; ++j) {
                if (num[j].l > num[i].l &&
                    num[j].w > num[i].w)
                    max_temp = max(max_temp, dp[j]);
            }
            dp[i] = max_temp + num[i].h;

            max_high = max(max_high, dp[i]);
        }

        printf("Case %d: maximum height = %d\n", ++t, max_high);
    }

    return 0;
}