HDU 1069 Monkey and Banana(DP 長方體堆放問題)

標籤：hdu   acm   dp   立方堆疊   

Monkey and BananaProblem 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

 題意   給你n種長方體 每種都有無窮個 三條棱長為a,b,c  當一個長方體的長寬都小於另一個時  這個長方體就可以堆在另一個上面  求這些長方體能堆起的最大高度

每個長方體都有6种放置方式  但只有三種高度  分別為a,b,c  為了便於操坐  可以把一個長方體分為三個 每個的高度都是唯一的  然後就可以用最長連通來求了  令d[i]表示以第i個長方體為最頂上一個時的最大高度    當第i個長方體的長和寬小於第j個的長和寬或寬和長時 第i個就可以放在第j個上面  即d[i]=max(d[i],d[j]+a[i].h)

#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int N = 35 * 3;int d[N], n;struct Cube{    int a, b, c;    Cube (int aa = 0, int bb = 0, int cc = 0) : a (aa), b (bb), c (cc) {}} cub[N];int dp (int i){    if (d[i] > 0) return d[i];    d[i] = cub[i].c;    for (int j = 1; j <= 3 * n; ++j)        if ( (cub[i].a < cub[j].a && cub[i].b < cub[j].b) || (cub[i].a < cub[j].b && cub[i].b < cub[j].a))            d[i] = max (d[i], dp (j) + cub[i].c);    return d[i];}int main(){    int cas = 0, ans, a, b, c;    while (scanf ("%d", &n), n)    {        memset (d, 0, sizeof (d));        for (int i = ans = 0; i < n; ++i)        {            scanf ("%d%d%d", &a, &b, &c);            cub[3 * i + 1] = Cube (a, b, c);            cub[3 * i + 2] = Cube (a, c, b);            cub[3 * i + 3] = Cube (b, c, a);        }        for (int i = 1; i <= 3 * n; ++i)            ans = max (ans, dp (i));        printf ("Case %d: maximum height = %d\n", ++cas, ans);    }    return 0;}