题目链接
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 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.
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”.
这题是很明显的求 DAG 上的最长路，也即矩形嵌套的变种题~ 因为每一种正方体是无限量的，所以这题要考虑自己是否可以放在自己的上面，那么就可以把每一个长方体拆成三个长方形，剩余的那一条边就作为最长路的距离，建好图后直接递归求最长路即可，AC代码如下：
#include<bits/stdc++.h> using namespace std; int n,cnt,ans,x,y,z,t=1,dp[100]; struct node{ int x,y,height; }block[100]; vector<int>g[100]; void build(){ for(int i=0;i<cnt;i++) g[i].clear(); for(int i=0;i<cnt;i++){ for(int j=0;j<cnt;j++){ if((block[i].x>block[j].x&&block[i].y>block[j].y)||(block[i].y>block[j].x&&block[i].x>block[j].y)) g[i].push_back(j); } } } int dfs(int u){ if(dp[u]) return dp[u]; dp[u]=block[u].height; for(int v:g[u]){ dp[u]=max(dp[u],dfs(v)+block[u].height); } return dp[u]; } int main(){ while(~scanf("%d",&n)&&n){ cnt=ans=0; for(int i=0;i<n;i++){ scanf("%d%d%d",&x,&y,&z); block[cnt].x=x,block[cnt].y=y,block[cnt++].height=z; block[cnt].x=x,block[cnt].y=z,block[cnt++].height=y; block[cnt].x=y,block[cnt].y=z,block[cnt++].height=x; } build(); memset(dp,0,sizeof(dp)); for(int i=0;i<cnt;i++){ if(!dp[i]) dfs(i); ans=max(ans,dp[i]); } printf("Case %d: maximum height = %d\n",t++,ans); } return 0; }