hdu 1069 Monkey and Banana(簡單dp)

果7發表於2014-01-22
NaN

Monkey and Banana

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6406    Accepted Submission(s): 3270


Problem 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





 








題目大意：主要是在讀題目時花了不少時間，就是給你n組x,y,z，然後x,y,z可以任意組合形成一個長方體的長寬高，那麼排列是有6種，然後我們可以將這6*n個長方體，選一些出來，使得立方體能夠一個挨一個從下面往上摞在一起，如果a摞在b上面，那麼需要ax<bx&&ay<by或者ax<by&&ay<bx，就是上面的可以摞在上面，然後求解這些長方體可以摞在一起的最大的高之和。





     解題思路：我們可以先選取一個作為高，然後把其它兩個分別作為長和寬，我們如果規定長>=寬，那麼每次擴張6種可以壓縮成為3種。n最大為30，可以先對這3n個立方體排序然後動態規劃即可。具體見程式碼。





     題目地址：Monkey and Banana





AC程式碼：


#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

struct node
{
    int x;
    int y;
    int z;
}nod[105];

int dp[105];

int cmp(node a,node b)
{
    if(a.x<b.x) return 1;
    if(a.x==b.x&&a.y<b.y) return 1;
    return 0;
}

int main()
{
    int n,cas=1;
    while(cin>>n&&n>0)
    {
        int i,j,m=0;
        int a[3];
        for(i=0;i<n;i++)
        {
            cin>>a[0]>>a[1]>>a[2];     //每種有6種情況，但是由於x1<x2&&y1<y2可以縮減為3種情況
            nod[m].x=max(a[0],a[1]),nod[m].y=min(a[0],a[1]),nod[m++].z=a[2];
            nod[m].x=max(a[1],a[2]),nod[m].y=min(a[1],a[2]),nod[m++].z=a[0];
            nod[m].x=max(a[0],a[2]),nod[m].y=min(a[0],a[2]),nod[m++].z=a[1];
        }

        sort(nod,nod+m,cmp);
        for(i=0;i<m;i++)
        {
            int ma=0;
            for(j=0;j<i;j++)
            {
                if(nod[j].x<nod[i].x&&nod[j].y<nod[i].y)
                    ma=max(ma,dp[j]);
            }
            dp[i]=ma+nod[i].z;
        }

        int res=0;
        for(i=0;i<m;i++)
            res=max(res,dp[i]);

        printf("Case %d: maximum height = %d\n",cas++,res);
    }
    return 0;
}

/*
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
*/






            




相關文章