POJ 3624 -- Charm Bracelet (0/1 Knapsack)


POJ 3624 -- Charm Bracelet (0/1 Knapsack)
Description
Bessie has gone to the mall's jewelry store and spies a charm bracelet. Of course, she'd like to fill it with the best charms possible from the N (1 ≤ N ≤ 3,402) available charms. Each charm i in the supplied list has a weight Wi (1 ≤ Wi ≤ 400), a 'desirability' factor Di (1 ≤ Di ≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more than M (1 ≤ M ≤ 12,880).
Given that weight limit as a constraint and a list of the charms with their weights and desirability rating, deduce the maximum possible sum of ratings.
Input
* Line 1: Two space-separated integers: N and M
* Lines 2..N+1: Line i+1 describes charm i with two space-separated integers: Wi and Di
Output
* Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints
Sample Input
4 6
1 4
2 6
3 12
2 7
Sample Output
23
http://blog.csdn.net/lihao21/article/details/6227445
Slyar.
Slyar:标准的01背包,动态转移方程如下。其中dp[i,j]表示的是前i个物品装入容量为j的背包里所能产生的最大价值,w[i]是第i个物品的重量,d[i]是第i个物品的价值。如果某物品超过背包容量,则该物品一定不放入背包,问题变为剩余i-1个物品装入容量为j的背包所能产生的最大价值;否则该物品装入背包,问题变为剩余i-1个物品装入容量为j-w[i]的背包所能产生的最大价值加上物品i的价值d[i]
  1. if (w[i]>j)   
  2.     dp[i,j] = dp[i-1,j]  
  3. else  
  4.     dp[i,j] = max(dp[i-1,j-w[i]]+d[i],dp[i-1,j])  
先上一个用二维数组求解的代码,这个代码比较容易理解。
  1. int main()  
  2. {  
  3.     int n, m;  
  4.     cin >> n >> m;  
  5.   
  6.     /*从下标1开始存放*/  
  7.     for(int i = 1; i <= n; i++) {  
  8.         cin >> w[i] >> d[i];  
  9.     }  
  10.   
  11.     /*注意,i==1或者j==1时,会用到dp边界下标为0的元素,由于已经初始化这些元素为0,保证了程序的正确运行*/  
  12.     for(int i = 1; i <= n; i++) {  
  13.         for(int j = 1; j <= m; j++) {  
  14.             if(w[i] <= j)  
  15.                 dp[i][j] = max(dp[i-1][j-w[i]] + d[i], dp[i-1][j]);  
  16.             else  
  17.                 dp[i][j] = dp[i-1][j];  
  18.         }  
  19.     }  
  20.   
  21.     cout << dp[n][m] << endl;  
  22.   
  23.     return 0;  
  24. }  
由于OJ上memory所限,上面代码提交上去会MLE的。在纸上计算dp过程中,发觉dp设置成n行的二维数组实际上没有必要,因为在计算dp第i行时只用到dp第i-1行的数据,而且到了最后,前面n-1行都没用的。因此,完全可以用一维数组进行计算。但要注意,应从后住前计算dp,否则dp的数据会被新计算的值覆盖,导致不正确。
  1.     for(int i = 1; i <= n; i++) {  
  2.         for(int j = m; j >= 1; j--) {  
  3.             if(w[i] <= j)  
  4.                 dp[j] = max(dp[j-w[i]] + d[i], dp[j]);  
  5.             else  
  6.                 dp[j] = dp[j];  
  7.         }  
  8.     }  
观察二重for循环,可以发觉其实还可以继续进一步优化下
  1.     for(int i = 1; i <= n; i++) {  
  2.         for(int j = m; j >= w[i]; j--) {  
  3.             dp[j] = max(dp[j-w[i]] + d[i], dp[j]);  
  4.         }  
  5.     }  
  6.   
Read full article from 3624 -- Charm Bracelet

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