Monkey and Banana

标签：状态   and   class   monkey   case   swap   VID   dimens   rect   

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  

题目大意：用给定的n个箱子（箱子可以重复使用，但是长和宽要严格递减），求可以构成的最大高度。
解题思路：贪心+dp，求出所有箱子可能的状态，对长和宽进行排序。

#include <bits/stdc++.h>
using namespace std;
int dp[105];
struct node
{
    int l,w,h;    
}t[105];//存放箱子 
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;
}
int main()
{
    int n,m,k,g=0;;
    while(cin>>n,n)
    {
        int a,b,c;
        k=0;
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++)
        {
            cin>>a>>b>>c;
            t[++k].l=a;t[k].w=b;t[k].h=c;
            if(t[k].l<t[k].w) swap(t[k].l,t[k].w);
            t[++k].l=b;t[k].w=c;t[k].h=a;
            if(t[k].l<t[k].w) swap(t[k].l,t[k].w);
            t[++k].l=c;t[k].w=a;t[k].h=b;
            if(t[k].l<t[k].w) swap(t[k].l,t[k].w);
        }//箱子的高度有三种放法 
        t[0].l=1e9,t[0].w=1e9,t[0].h=0;
        sort(t+1,t+1+k,cmp);
        int maxx=0;
        for(int i=1;i<=k;i++)
        {
            for(int j=0;j<i;j++)
            {
                if(t[i].l<t[j].l&&t[i].w<t[j].w)
                {
                    dp[i]=max(dp[i],dp[j]+t[i].h);    
                    maxx=max(maxx,dp[i]); 
                }    
            }
        }//下降子序列 
        g++;
        printf("Case %d: maximum height = %d\n",g,maxx);
        
    }    
} 

Monkey and Banana

标签：状态   and   class   monkey   case   swap   VID   dimens   rect   

原文地址：https://www.cnblogs.com/ww123/p/10651734.html