UVa 437 The Tower of Babylon & NYOJ 232 How to eat more Banana

时间限制：1000 ms  |  内存限制：65535 KB

难度：4

描述

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.

输入

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.

输出

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".

样例输入

110 20 3026 8 105 5 571 1 12 2 23 3 34 4 45 5 56 6 67 7 7531 41 5926 53 5897 93 2384 62 6433 83 270

样例输出

Case 1: maximum height = 40Case 2: maximum height = 21Case 3: maximum height = 28Case 4: maximum height = 342

DAG上的动态规划，先将x、y、z按从小到大排序，每个长方体按x，y，z；x，z，y；y，z，x存三次，分别以a[i].x，a[i].y，a[i].z为第一、二、三关键字从小到大排序。更新满足条件的dp[i]。（注：dp[i]=max(dp[i],dp[j]+a[j].z)不是状态转移方程，只是一个更新公式，不能直接算出dp[i]）代码如下：

#include<iostream>#include<algorithm>using namespace std;typedef struct node{    int x,y,z;}node;bool cmp(const node a,const node b){    if(a.x==b.x)    {        if(a.y==b.y)        {            return a.z<b.z;        }        else return a.y<b.y;    }    else return a.x<b.x;}node a[100];int main(){    int t,cou=0;    while(cin>>t&&t)    {        t*=3;        int ans=0;        for(int i=0;i<t;i+=3)        {            int x,y,z;            cin>>x>>y>>z;            if(x>y) swap(x,y);            if(y>z) swap(y,z);            if(x>y) swap(x,y);            a[i].x=x,a[i].y=y,a[i].z=z;            a[i+1].x=x,a[i+1].y=z,a[i+1].z=y;            a[i+2].x=y,a[i+2].y=z,a[i+2].z=x;        }        sort(a,a+t,cmp);        int dp[100];        for(int i=0;i<t;++i) dp[i]=a[i].z;        for(int i=0;i<t;++i)        {            for(int j=i;j>=0;--j)            {                if(a[i].x>a[j].x&&a[i].y>a[j].y)                {                    dp[i]=max(dp[i],dp[j]+a[i].z);                }                ans=max(dp[i],ans);            }        }        cout<<"Case "<<++cou<<": maximum height = "<<ans<<endl;    }    return 0;}