ボックススタッキングの問題

この問題では、さまざまなボックスのセットが提供されます。長さ、幅、幅はボックスごとに異なる場合があります。私たちの仕事は、これらの箱の積み重ねを見つけることです。その高さは可能な限り高くなっています。好きなようにボックスを回転させることができます。ただし、維持するルールがあります。

下のボックスの上面の面積が上のボックスの下部の面積よりも大きい場合は、別のボックスにボックスを配置できます。

入力と出力

Input:
A list of boxes is given. Each box is denoted by (length, bredth, height).
{ (4, 6, 7), (1, 2, 3), (4, 5, 6), (10, 12, 32) }
Output:
The maximum possible height of box stack is: 60

アルゴリズム

maxHeight(boxList, n)

入力- さまざまなボックスのリスト、ボックスの数。

出力- ボックスを積み重ねて見つけた最大の高さ。

Begin
   define rotation array rot of size 3n.
   index := 0

   for all boxes i, in the boxList, do
      rot[index].len := boxList[i].len
      rot[index].hei := maximum of boxList[i].hei and boxList[i].bre
      rot[index].bre := minimum of boxList[i].hei and boxList[i].bre
      index := index + 1

      rot[index].len := boxList[i].bre
      rot[index].hei := maximum of boxList[i].len and boxList[i].hei
      rot[index].bre := minimum of boxList[i].len and boxList[i].hei
      index := index + 1

      rot[index].len := boxList[i].hei
      rot[index].hei := maximum of boxList[i].len and boxList[i].bre
      rot[index].bre := minimum of boxList[i].len and boxList[i].bre
      index := index + 1

      n := 3n
      sort the rot list
      define maxHeightTemp array

      for i := 1 to n-1, do
         for j := 0 to i-1, do
            if rot[i].bre < rot[j].bre AND
               rot[i].hei < rot[j].hei AND
               maxHeightTemp[i] < maxHeightTemp + rot[i].len, then
               maxHeightTemp[i] := maxHeightTemp[j] +
               rot[i].len
         done
      done

      maxHeight := -1
      for i := 0 to n-1, do
         if maxHeight < maxHeightTemp[i], then
            maxHeight := maxHeightTemp[i]
      done
   done
   return maxHeight
End

例

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

struct Box {
   int length, bredth, height;
};

int min (int x, int y) {
   return (x < y)? x : y;
}

int max (int x, int y) {
   return (x > y)? x : y;
}

bool compare(Box b1, Box b2) {
   return b1.height > b2.height;    //to sort the box as descending order of height
}

int maxHeight( Box boxList[], int n ) {
   Box rotation[3*n];    //a box can be rotared as 3 type, so there is 3n number of rotations
   int index = 0;

   for (int i = 0; i < n; i++) {
      //store initial position of the box
      rotation[index].length = boxList[i].length;
      rotation[index].height = max(boxList[i].height, boxList[i].bredth);
      rotation[index].bredth = min(boxList[i].height, boxList[i].bredth);
      index++;

      //dimention after first rotation
      rotation[index].length = boxList[i].bredth;
      rotation[index].height = max(boxList[i].length, boxList[i].height);
      rotation[index].bredth = min(boxList[i].length, boxList[i].height);
      index++;        
   
      //Dimention after second rotation
      rotation[index].length = boxList[i].height;
      rotation[index].height = max(boxList[i].length, boxList[i].bredth);
      rotation[index].bredth = min(boxList[i].length, boxList[i].bredth);
      index++;
   }

   n = 3*n;    //set n as 3n for 3 rotations of each boxes

   sort(rotation, rotation+n, compare);    //sort rotation array as descending order

   int maxHTemp[n];    //temporary max height if ith box is stacked

   for (int i = 0; i < n; i++ )
      maxHTemp[i] = rotation[i].length;
   
   for (int i = 1; i < n; i++ )    //find optimized stack height
      for (int j = 0; j < i; j++ )
         if ( rotation[i].bredth < rotation[j].bredth && rotation[i].height < rotation[j].height
            && maxHTemp[i] < maxHTemp[j] + rotation[i].length) {
            maxHTemp[i] = maxHTemp[j] + rotation[i].length;
         }
   int maxHeight = -1;
   for ( int i = 0; i < n; i++ )    //find the maximum height from all temporary heights
         
      if ( maxHeight < maxHTemp[i] )
         maxHeight = maxHTemp[i];
         
   return maxHeight;
}

int main() {
   Box arr[] = { {4, 6, 7}, {1, 2, 3}, {4, 5, 6}, {10, 12, 32} };
   int n = 4;
   cout<<"The maximum possible height of box stack is: " << maxHeight (arr, n) << endl;
}

出力

The maximum possible height of box stack is: 60