当前位置： 移动技术网 > IT编程>开发语言>C/C++ > C - Monkey and Banana HDU 1069( 动态规划+叠放长方体)

C - Monkey and Banana HDU 1069( 动态规划+叠放长方体)

2018年12月10日  | 移动技术网IT编程  | 我要评论

我相信伴奏,疯果网,安全教育手抄报

c - monkey and banana

time limit:1000ms memory limit:32768kb 64bit io format:%i64d & %i64u

submit status practice hdu 1069

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.


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

 

刚开始学动态规划，还是挺难的，我参考了这位大神的思路

dp[i]:表示以p[i]这个长方体放在顶端，可以获得最大的高度。

那么状态转移是，dp[i]=max( dp[i], p[i].high + dp[j] )

 


#include
#include
#include
#include
#include
#include
using namespace std;
templateinline t read(t&x)
{
    char c;
    while((c=getchar())<=32)if(c==eof)return 0;
    bool ok=false;
    if(c=='-')ok=true,c=getchar();
    for(x=0; c>32; c=getchar())
        x=x*10+c-'0';
    if(ok)x=-x;
    return 1;
}
template inline t read_(t&x,t&y)
{
    return read(x)&&read(y);
}
template inline t read__(t&x,t&y,t&z)
{
    return read(x)&&read(y)&&read(z);
}
template inline void write(t x)
{
    if(x<0)putchar('-'),x=-x;
    if(x<10)putchar(x+'0');
    else write(x/10),putchar(x%10+'0');
}
templateinline void writeln(t x)
{
    write(x);
    putchar('\n');
}
//-------zcc io template------
const int maxn=400;
const double inf=999999999;
#define lson (rt<<1),l,m
#define rson (rt<<1|1),m+1,r
#define m ((l+r)>>1)
#define for(i,t,n) for(int i=(t);i<(n);i++)
typedef long long  ll;
typedef double db;
typedef pair p;
#define bug printf("---\n");
#define mod  1000000007

struct node
{
    int width,length,area,high;
    bool operator<(const node&a) const
    {
        return area>a.area;
    }
}p[maxn];
int dp[maxn];

int main()
{
    int n,m,i,j,t,k;
    int cas=1;
    while(read(n)&&n)
    {
        k=0;
        for(i,0,n)
        {
            int w,l,h;
            read__(w,l,h);
            node tmp;
            tmp.width=w;tmp.length=l;tmp.high=h;
            tmp.area=w*l;
            p[k++]=tmp;
            tmp.width=w;tmp.length=h;tmp.high=l;
            tmp.area=w*h;
            p[k++]=tmp;
            tmp.width=h;tmp.length=l;tmp.high=w;
            tmp.area=h*l;
            p[k++]=tmp;
        }
        sort(p,p+k);
        dp[0]=p[0].high;
        for(i=1;ip[i].width&&p[j].length>p[i].length||p[j].width>p[i].length&&p[j].length>p[i].width)
            {
                dp[i]=max(dp[i],dp[j]+p[i].high);
            }
        }
//        for(i,0,k)
//        printf("dp[%d] = %d\n",i,dp[i]);
        printf("case %d: maximum height = %d\n",cas++,*max_element(dp,dp+k));
    }
    return 0;
}

 

您可能感兴趣的文章:

如对本文有疑问，请在下面进行留言讨论，广大热心网友会与你互动！！ 点击进行留言回复

相关文章:

验证码：
最近更新的文章
大家感兴趣的文章
移动技术网