HDU 1069(动态规划)

清疚 2022-06-06 03:15 152阅读 0赞

问题描述:

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.

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

题目题意:题目给了我们n种石头（矩形），每种石头无限，石头可以任意摆放，问石头最高可以摆多高？上面的石头必须严格大于下面的石头！

题目分析:我们对每种石头的六种摆放形式都枚举出来，然后按照面积小排序！然后DP了 dp\[i\]表示以石头i结束的最高高度！

代码如下:

#include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<cmath>
    #include<algorithm>
    using namespace std;
    
    const int maxn=200;
    struct note
    {
        int x,y,h;
    }point[maxn],dp[maxn];
    
    
    bool cmp(note a,note b)
    {
        if (a.x*a.y==b.x*b.y) return a.h<b.h;
        else return a.x*a.y<b.x*b.y;
    }
    int main()
    {
        int n,res,ans,icase=0;
        while (scanf("%d",&n)!=EOF) {
            res=ans=0;
            if (n==0) break;
            for (int i=0;i<n;i++) {
                int a,b,c;
                scanf("%d%d%d",&a,&b,&c);//枚举
                point[res].x=a,point[res].y=b,point[res++].h=c;
                point[res].x=b,point[res].y=a,point[res++].h=c;
                point[res].x=a,point[res].y=c,point[res++].h=b;
                point[res].x=c,point[res].y=a,point[res++].h=b;
                point[res].x=b,point[res].y=c,point[res++].h=a;
                point[res].x=c,point[res].y=b,point[res++].h=a;
            }
            sort (point,point+res,cmp);
            memset (dp,0,sizeof (dp));
            for (int i=0;i<res;i++) {
                dp[i]=point[i];
                for (int j=0;j<i;j++) {
                    if (dp[j].x<dp[i].x&&dp[j].y<dp[i].y||dp[j].y<dp[i].x&&dp[j].x<dp[i].y)//严格大于
                        dp[i].h=max(dp[i].h,point[i].h+dp[j].h);//转移方程
                }
                ans=max(ans,dp[i].h);
            }
            printf("Case %d: maximum height = %d\n",++icase,ans);
    
        }
        return 0;
    }