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法律研究箱子堆叠问题
在这个问题中给定了一组不同的盒子，不同盒子的长度、宽度和高度可能不同。我们的任务是找到一叠这些盒子的高度尽可能大。我们可以根据需要旋转任何盒子。但有一条规则需要遵守。
只有当底部盒子的顶部面积大于顶部盒子的底部面积时，才能将一个盒子放在另一个盒子上。
输入和输出
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算法
maxHeight(boxList, n)
输入 − 不同盒子列表、盒子数。
输出 −通过堆叠盒子找到的最大高度。
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
示例
#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;
}输出
The maximum possible height of box stack is: 60
更新时间：16-Jun-2020
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