HDU 1069 Monkey and Banana (動規)

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Monkey and Banana Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7248    Accepted Submission(s): 3730


Problem DescriptionA 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.
 
InputThe 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.
 
OutputFor 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

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

 
Sample Output

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

 
SourceUniversity of Ulm Local Contest 1996 
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題意：

有一堆箱子，有長寬高，x,y,z。  規定：放在上面的箱子，無論長和寬都要比下面的箱子大，只有一邊大是不行的。

箱子的方向可以任意放。 這樣，x,y,z就有六種組合。其實也可以說是三種(如果要的是面積的大小)。

因為要和上面的箱子比較，這裡為了計算方便，直接寫成六種。 

求的是最大高度。

注意：不是只有一個箱子，而是很多那種規格的箱子，開始以為只有一個，第一個範例就不能解釋。

只不過這個情況，我們需要先排下序。把大的箱子放前面。有點牽強的說，也算是用了貪心的思想。

雖然我們不能確定一個x邊最大的箱子不一定就能放在最下面，例如的它的y邊很小，但是，它肯定也放不到其它箱子的上面，所以，把它放前面也是可以的。

因此可以按x邊排序。不過在最後比較的時候，要嚴格遵循 x1 > x2 && y1 > y2。

#include <iostream>#include <algorithm>using namespace std;#define M 1000struct node{    int x,y,z;}vis[M];int mat[M];bool cmp(node a,node b){    if(a.x==b.x)        return a.y>b.y;    return a.x>b.x;}int main(int a,int b,int c){    int i,j,n,k=0;    while(scanf("%d",&n)!=EOF&&n)    {   memset(mat,0,sizeof(mat));        k++;        for(i=0;i<n*6;i+=6)        {   //箱子的不同放法。            scanf("%d%d%d",&a,&b,&c);            {                vis[i].x=a;                vis[i].y=b;                vis[i].z=c;                                vis[i+1].x=a;                vis[i+1].y=c;                vis[i+1].z=b;                                vis[i+2].x=b;                vis[i+2].y=a;                vis[i+2].z=c;                                vis[i+3].x=b;                vis[i+3].y=c;                vis[i+3].z=a;                                vis[i+4].x=c;                vis[i+4].y=a;                vis[i+4].z=b;                                vis[i+5].x=c;                vis[i+5].y=b;                vis[i+5].z=a;            }        }            sort(vis,vis+n*6,cmp); //排序。            for(i=0;i<n*6;i++)            {                mat[i]=vis[i].z;     //初始情況：只有自己，下面沒有墊箱子。                for(j=i-1;j>=0;j--)  //從i->0和0->i都可以，因為mat[j]就是第j個箱子在最上面的最優解，這個在迴圈裡已經求過。                {                    if(vis[i].x<vis[j].x&&vis[i].y<vis[j].y)                        if(mat[i]<mat[j]+vis[i].z)           //找到一個更優解，替換掉原來的解。                        {                            mat[i]=mat[j]+vis[i].z;                        }                }            }            int max=0;            for(i=0;i<n*6;i++)            if(max<mat[i]) max=mat[i];             //尋找每一個箱子在最上面的時候的高度最大值。            printf("Case %d: maximum height = %d\n",k,max);    }    return 0;}