Monkey and Banana

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16589 Accepted Submission(s): 8834

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

Source

University of Ulm Local Contest 1996

O(n^3)：

1. 设dp[i][j]为第i块放块j的最大高度。

2.状态转移：对于第i块放块j的情况下，枚举所有块k，如果块j能放在块k上面，那么第i块放块j 就可以从第i-1块放块k 中转移过来了。

即： dp[i][j] = max(dp[i-1][k]+height[j])

代码如下：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int MAXN = 100;
 4 
 5 int block[MAXN][3];
 6 int dp[MAXN][MAXN];
 7 
 8 int main()
 9 {
10     int n, kase = 0;
11     while(scanf("%d",&n) && n)
12     {
13         int N = 0;
14         for(int i = 1; i<=n; i++)
15         {
16             int a, b, h;
17             scanf("%d%d%d", &a, &b, &h);    //三种放置如下：
18             block[++N][0] = min(b,h), block[N][1] = max(b,h), block[N][2] = a;
19             block[++N][0] = min(a,h), block[N][1] = max(a,h), block[N][2] = b;
20             block[++N][0] = min(a,b), block[N][1] = max(a,b), block[N][2] = h;
21         }
22 
23         int ans = -2000000;
24         memset(dp, 0, sizeof(dp));
25         for(int i = 1; i<=N; i++)   //初始化第一个
26             dp[1][i] = block[i][2], ans = max(dp[1][i], ans);
27 
28         for(int i = 2; i<=N; i++)   //第i个
29         for(int j = 1; j<=N; j++)   //第i个为块j
30         for(int k = 1; k<=N; k++)   //枚举块j下面的块
31              if(block[j][0]<block[k][0] && block[j][1]<block[k][1]) //块j能够放在块k上， 那么就可以转移
32                 dp[i][j] = max(dp[i][j], dp[i-1][k]+block[j][2]), ans = max(dp[i][j], ans);
33 
34         printf("Case %d: maximum height = %d
", ++kase, ans);
35     }
36     return 0;
37  }

View Code

O(n^2)：

1.根据长或者宽，对每一种block（每一块block有三种放置方式）进行降序排序。

2.设dp[i]为块i放在最上面的最大高度。

3.对于当前块i，枚举能够放在它下面的块j（由于经过了排序，所以j的下标为1~i-1），然后把块i放到块j上，更新dp[i]。思想与LIS的O(n^2)写法类似。

4.相同类型的题：HDU1160

代码如下：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int MAXN = 100;
 4 
 5 struct node
 6 {
 7     int a, b, h;
 8     bool operator<(const node x){   //对a或者b进行排序（降序）
 9         return a>x.a;
10     }
11 }block[MAXN];
12 int dp[MAXN];
13 
14 int main()
15 {
16     int n, kase = 0;
17     while(scanf("%d",&n) && n)
18     {
19         int N = 0;
20         for(int i = 1; i<=n; i++)
21         {
22             int a, b, h;
23             scanf("%d%d%d", &a, &b, &h);    //三种放置如下：
24             block[++N].a = min(b,h), block[N].b = max(b,h), block[N].h = a;
25             block[++N].a = min(a,h), block[N].b = max(a,h), block[N].h = b;
26             block[++N].a = min(a,b), block[N].b = max(a,b), block[N].h = h;
27         }
28         sort(block+1, block+1+N);
29 
30         int ans = -2000000;
31         for(int i = 1; i<=N; i++)    //初始化第一个
32             dp[i] = block[i].h, ans = max(ans, dp[i]);
33 
34         //对于当前i，枚举能够放在它下面的块j，然后跟新dp[i]。
35         for(int i = 1; i<=N; i++)
36             for(int j = 1; j<i; j++)
37                 if(block[i].a<block[j].a && block[i].b<block[j].b)
38                     dp[i] = max(dp[i], dp[j]+block[i].h), ans = max(ans, dp[i]);
39 
40         printf("Case %d: maximum height = %d
", ++kase, ans);
41     }
42     return 0;
43  }

View Code