买书问题——编程之美1.4

//买书问题，
//如果一种书籍五册，单独买一册8元，买两册不同的打95折，买三册不同的书籍9折，买四册不同的书籍8折，买五册不同书籍75折，问怎么买书最便宜。
//比如买2本一册，2本2册，2本三册，1本4册，1本5册
//那么最优打折方式就是：分两次购买，一本一册，一本二册，一本三册和一本四册，然后就是剩下的书籍
//此问题用到动态规划，
#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;
//默认x1>x2>x3>x4>x5;
double money(int x1,int x2,int x3,int x4,int x5){
	if(x1 < x2 || x2 < x3 || x3 < x4 || x4 < x5){//保证满足我们的默认情况
		vector<int> tmp;
		tmp.push_back(x1);
		tmp.push_back(x2);
		tmp.push_back(x3);
		tmp.push_back(x4);
		tmp.push_back(x5);
		sort(tmp.begin(),tmp.end());
		x1 = tmp[4];
		x2 = tmp[3];
		x3 = tmp[2];
		x4 = tmp[1];
		x5 = tmp[0];
		money(x1,x2,x3,x4,x5);
	}
	if(x1 == 0)
		return 0;
	double res1,res2,res3,res4,res5;
	res1 = res2 = res3 = res4 = res5 = 0;
	if(x5 >= 1)
		res5 = 5 * 8 * (1 - 0.25) + money(x1 - 1, x2 - 1,x3 - 1, x4 - 1, x5 - 1);
	if(x4 >= 1)
		res4 = 4 * 8 * (1 - 0.2) + money(x1 - 1, x2 - 1,x3 - 1, x4 - 1, x5);
	if(x3 >= 1)
		res3 = 3 * 8 * (1 - 0.1) + money(x1 - 1, x2 - 1,x3 - 1, x4, x5);
	if(x2 >= 1)
		res2 = 2 * 8 * (1 - 0.05) + money(x1 - 1, x2 - 1,x3, x4, x5);
	if(x1 >= 1)
		res1 = 8 + money(x1 - 1, x2, x3, x4, x5);
	if(x2 == 0)
		return res1;
	if(x3 == 0)
		return min(res1,res2);
	if(x4 == 0)
		return min(res1,min(res2,res3));
	if(x5 == 0)
		return min(res1,min(res2,min(res3,res4)));
	if(x5 > 0)
		return min(res1,min(res2,min(res3,min(res4,res5))));
}


int main()
{
	cout<<money(2,2,2,1,1);
	cin.get();
	return 0;
}