Q9.10 To build the tallest stack

Q:You have a stack of n boxes, with widths w., heights l\ and depths dr The boxes cannot be rotated and can only be stacked on top of one another if each box in the stack is strictly larger than the box above it in width, height, and depth. Implement a method to build the tallest stack possible, where the heigh t of a stack is the sum of the heights of each box.

A: 给定箱子a,b,c,d,e...，那么结果为max(以a为底的栈，以b为底的栈，以c为底的栈，以d为底的栈，以e为底的栈...)。 子问题用同样的方法求解，只是用最底层的判断转为对倒数第二层的判断。为了节省时间，采用备忘录方法。

#include <iostream>#include <map>#include <vector>#include <cstdio>using namespace std;class box {private:float h, w, d;public:box(float H, float W, float D){h = H;w = W;d =D;}bool canBeAbove(box B);float getDepth();};bool box::canBeAbove(box B){if (h<B.h && w<B.w && d<B.d)return true;return false;}float box::getDepth(){return d;}typedef vector<box> vb;typedef map<int, vb> mvb;float stackHeight(vb Boxes){float height = 0;for (int i = 0; i < Boxes.size(); ++i) {height += Boxes[i].getDepth();}return height;}vb createStack(mvb &Map, vb Boxes, int bottom) {if (Map.count(bottom)!= 0 && bottom < Boxes.size()) {return Map[bottom];}int max_height = 0;vb max_stack;if (bottom >= Boxes.size()) {return max_stack;}for (int i = 0; i < Boxes.size(); ++i) {if (Boxes[i].canBeAbove(Boxes[bottom])) {vb tmpStack = createStack(Map, Boxes, i);int tmpHeight = stackHeight(tmpStack);if (tmpHeight > max_height) {max_stack = tmpStack;max_height = tmpHeight;}}}//Push Boxes[bottom] to font heemax_stack.push_back(Boxes[bottom]);Map[bottom] = max_stack;return max_stack;}int main(){freopen("Question9_10.in", "r", stdin);int n;cin>>n;mvb Map;vb Boxes, result;float H, W, D;for (int i = 0; i < n; ++i) {cin>>H>>W>>D;cout<<H<<" "<<W<<" "<<D<<endl;Boxes.push_back(box(H, W, D));}for (int i = 0; i < n; i++) {vb tmp = createStack(Map, Boxes, i);if (result.size() < tmp.size()) {result = tmp;}}for (int i = 0; i < result.size(); ++i) {cout<<result[i].getDepth()<<" ";}return 0;}

其中输入数据为：

4111333484222