买书问题 - 编程之美1.4


买书问题 - 编程之美1.4
上柜的《哈利波特》平装本系列,一共有五卷。假设每一卷单独销售均需8欧元。如果读者一次购买不同的两卷,就可以扣除5%的费用,三卷则更多。假设具体折扣的情况如下:
        本数    2       折扣   5%
        本数    3       折扣  10%
        本数    4       折扣  20%
        本数    5       折扣  25% 
买书的需求一定,但通过不同的买书组合,能买到不同的折扣。
问题:设计出算法,能够计算出读者所购买的一批书的最低价格。
动态规划
        五卷书的价格相同都是8欧元,所以购买(1,0,0,0,0)跟(0,1,0,0,0)效果一样。这里就可以简化为,让所购买书按照本书递增(递减),从而方便讨论。
         要处理的参数为购买每种卷的个数,所以递归一定跟这五个参数相关。可以把参数按照从小到大顺序排列。讨论不为0的参数的个数,从而求出所有可能的折扣种类。然后从当前折扣种类中取价格最小值。
状态转移方程: 
                           (X1,X2,X3,X4,X5)代表购买每卷的个数,F(X1X2X3X4X5)代表最低价格。其中,X1 < X2 < X3 < X4 < X5
                            F(X1X2X3X4X5)=0  ;当所有参数都为0的情况(这也是退出递归的出口)
              F(X1X2X3X4X5)= min{
                    5*8*(1-25%) +F(X1-1,X2-1X3-1X4-1X5-1) //参数全部  > 0
                    4*8*(1-20%) +F(X1,X2-1X3-1X4-1X5-1)    //x2 > 0
                    3*8*(1-10%) +F(X1,X2X3-1X4-1X5-1)       //x3 > 0
                    2*8*(1-5%) +F(X1,X2X3X4-1X5-1)            //x4 > 0
                   8 +F(X1,X2X3X4X5-1)                                 //x5 > 0
             }
  1. double BestBuy(int x1, int x2, int x3, int x4, int x5)  
  2. {  
  3.     countRE++;  
  4.     int i;  
  5.     int n[5] ={ x1, x2, x3, x4, x5 };  
  6.     reRank(n,5);  
  7.     x1=n[0];  
  8.     x2=n[1];  
  9.     x3=n[2];  
  10.     x4=n[3];  
  11.     x5=n[4];  
  12.     if(S[x1][x2][x3][x4][x5]>0) return S[x1][x2][x3][x4][x5];  
  13.     countDP++;  
  14.     /* x1 < x2 < x3 < x4  < x5*/   
  15.     if (n[0]>0)  
  16.     {  
  17.         double solution = findMin( 8+BestBuy(x1, x2, x3, x4, x5 - 1),  
  18.             2 * 8 * 0.95 + BestBuy(x1, x2, x3, x4 - 1, x5 - 1),  
  19.             3 * 8 *0.9 + BestBuy(x1, x2, x3-1, x4 - 1, x5 - 1),  
  20.             4 * 8 * 0.8 + BestBuy(x1, x2 - 1, x3 - 1, x4 - 1, x5 - 1),   
  21.             5 * 8 * 0.75 + BestBuy(x1-1, x2 - 1, x3 - 1, x4 - 1, x5 - 1) );  
  22.         S[x1][x2][x3][x4][x5] = solution;  
  23.         return solution;  
  24.     }  
  25.     else if ((n[0] == 0) && (n[1] > 0))   
  26.     {  
  27.         double solution = findMin( 8+BestBuy(x1, x2, x3, x4, x5 - 1),  
  28.             2 * 8 * 0.95 + BestBuy(x1, x2, x3, x4 - 1, x5 - 1),  
  29.             3 * 8 *0.9 + BestBuy(x1, x2, x3-1, x4 - 1, x5 - 1),  
  30.             4 * 8 * 0.8 + BestBuy(x1, x2 - 1, x3 - 1, x4 - 1, x5 - 1), large );  
  31.         S[x1][x2][x3][x4][x5] = solution;  
  32.         return solution;  
  33.     }  
  34.     else if ((n[0] == 0) && (n[1] == 0) &&(n[2]>0))   
  35.     {  
  36.         double solution = findMin( 8+BestBuy(x1, x2, x3, x4, x5 - 1),  
  37.             2 * 8 * 0.95 + BestBuy(x1, x2, x3, x4 - 1, x5 - 1),  
  38.             3 * 8 *0.9 + BestBuy(x1, x2, x3-1, x4 - 1, x5 - 1), large, large );  
  39.         S[x1][x2][x3][x4][x5] = solution;  
  40.         return solution;  
  41.     }  
  42.     else if ((n[0] == 0) && (n[1] == 0) && (n[2] == 0) && (n[3] > 0))   
  43.     {  
  44.         double solution = findMin( 8+BestBuy(x1, x2, x3, x4, x5 - 1),  
  45.             2 * 8 * 0.95 + BestBuy(x1, x2, x3, x4 - 1, x5 - 1), large, large, large );  
  46.         S[x1][x2][x3][x4][x5] = solution;  
  47.         return solution;  
  48.     }  
  49.     else if ((n[0] == 0) && (n[1] == 0) && (n[2] == 0) && (n[3] == 0) && (n[4] > 0))   
  50.     {  
  51.         double solution = 8.0 + BestBuy(x1, x2, x3, x4, x5 - 1);   
  52.         S[x1][x2][x3][x4][x5] = solution;  
  53.         return solution;  
  54.     }  
  55.     else  
  56.     {  
  57.         S[x1][x2][x3][x4][x5] =0;  
  58.         return 0;  
  59.     }  
  1.     cout<<BestBuy(a[0],a[1],a[2],a[3],a[4])<<endl;  
《编程之美》读书笔记(四):卖书折扣问题的贪心解法
1 本次选择的书的数量不应该小于下一次可选择的书的最大数量(其中,书的取法按照书中动态规划方法给出的选法来进行)。
2 每一次选择之前都应该查表,选择其中使得近两次折扣数最大的那个作为本次选择(因为我们使得本次选择的影响被控制在两次选择的范围内)
  1.     public double findMinCost(int Y1,int Y2,int Y3,int Y4,int Y5){  
  2.         int[] x={Y1,Y2,Y3,Y4,Y5};  
  3.         Arrays.sort(x);  
  4.         if(x[0]<0){  
  5.             System.out.println("购书数量不能为负数");  
  6.             return 0;  
  7.         }  
  8.         //Y1>=Y2>=Y3>=Y4>=Y5  
  9.         Y5=x[0];  
  10.         Y4=x[1];  
  11.         Y3=x[2];  
  12.         Y2=x[3];  
  13.         Y1=x[4];  
  14.         double cost=0;  
  15.         if(Y5>=1){  
  16.             cost=min(  
  17.                     8*5*(1-0.25)+findMinCost(Y1-1,Y2-1,Y3-1,Y4-1,Y5-1),  
  18.                     8*4*(1-0.20)+findMinCost(Y1-1,Y2-1,Y3-1,Y4-1,Y5),  
  19.                     8*3*(1-0.10)+findMinCost(Y1-1,Y2-1,Y3-1,Y4,Y5),  
  20.                     8*2*(1-0.05)+findMinCost(Y1-1,Y2-1,Y3,Y4,Y5)  
  21.                 );  
  22.         }else if(Y4>=1){  
  23.             cost=min(  
  24.                     8*4*(1-0.20)+findMinCost(Y1-1,Y2-1,Y3-1,Y4-1,Y5),  
  25.                     8*3*(1-0.10)+findMinCost(Y1-1,Y2-1,Y3-1,Y4,Y5),  
  26.                     8*2*(1-0.05)+findMinCost(Y1-1,Y2-1,Y3,Y4,Y5)  
  27.                 );  
  28.         }else if(Y3>=1){  
  29.             cost=min(  
  30.                     8*3*(1-0.10)+findMinCost(Y1-1,Y2-1,Y3-1,Y4,Y5),  
  31.                     8*2*(1-0.05)+findMinCost(Y1-1,Y2-1,Y3,Y4,Y5)  
  32.                 );  
  33.         }else if(Y2>=1){  
  34.             cost=min(  
  35.                     8*2*(1-0.05)+findMinCost(Y1-1,Y2-1,Y3,Y4,Y5)  
  36.                 );  
  37.         }else if(Y1>=1){//{Y1,0,0,0,0},说明买的是同一卷书,没有折扣  
  38.             cost=8*Y1;  
  39.         }else{//{0,0,0,0,0}  
  40.             cost=0;  
  41.         }  
  42.         return cost;  
  43.     }  
Read full article from 每日一题(18)——买书问题(动态规划) - 小熊不去实验室 - 博客频道 - CSDN.NET

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