hdu 1069 Monkey and Banana
Monkey and Banana
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12865 Accepted Submission(s): 6757
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.
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.
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
Recommend
#include<iostream> #include<stdio.h> #include<algorithm> using namespace std; const int maxn = 30*4; int num=0; struct Node { int len,wid; int hei; } rect[maxn]; int dp[maxn]; void add(int t[],int sta) { if(sta==0) { rect[num].len=max(t[1],t[2]); rect[num].wid= min(t[1],t[2]); rect[num].hei=t[0]; num++; return; } if(sta==1) { rect[num].len=max(t[0],t[2]); rect[num].wid= min(t[0],t[2]); rect[num].hei=t[1]; num++; return; } rect[num].len=max(t[0],t[1]); rect[num].wid= min(t[0],t[1]); rect[num].hei=t[2]; num++; } bool cmp(Node a,Node b) { if(a.len!=b.len) return a.len<b.len; if(a.wid!=b.wid) return a.wid<b.wid; return a.hei>b.hei; } int main() { int n; int cas=1; while(~scanf("%d",&n)&&n) { num=0; for(int i=0;i<n;i++) { int t[4]; for(int k=0;k<3;k++) scanf("%d",t+k); for(int k=0;k<3;k++) add(t,k); } int maxx = -1; sort(rect,rect+num,cmp); for(int i=0;i<num;i++) dp[i]=rect[i].hei; for(int i=1;i<num;i++) { for(int j=i-1;j>=0;j--) { if(rect[i].len>rect[j].len&&rect[i].wid>rect[j].wid) { dp[i]=max(dp[i],dp[j]+rect[i].hei); } } maxx=max(dp[i],maxx); } printf("Case %d: maximum height = %d\n",cas++,maxx); } return 0; }
dp[i]表示 以i为最下面的箱子，可以得到的最大高度。
在判断第i的时候，由于箱子已经按照 base dimensions （也就是长和宽）从小到大递增，所以只能把第i个箱子放在前i-1个箱子的某一个下面
（这里便是dp数组循环所做的事情）
这样，从第一个箱子（以为第一个箱子上面不肯能在放其他的箱子，所以我们直接从第二个箱子开始）循环，每次记录最大值，
最后按照样例格式输出即可。
