文章作者：姜南(Slyar) 文章来源：Slyar Home (www.slyar.com) 转载请注明，谢谢合作。
DP 01背包
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
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]...恩，多清晰...
还有一点是，这题我之前使用二维数组做的，不过提交之后MLE了...所以只能想办法使用一维数组，手工模拟一下发现dp[i,j]只跟之前的某几个值有关，再之前计算的结果就都没用了...所以我们就可以用一个数组来重复使用了，不过注意要倒序，保证无后效性...
if (w[i]>j) dp[i,j] = dp[i-1,j]
else dp[i,j] = max(dp[i-1,j-w[i]]+d[i],dp[i-1,j])
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#include <iostream> using namespace std; const int N = 3403; const int M = 12881; int dp[M] = {0}; int max(int x, int y) { return x > y ? x : y; } int main() { int n, m; int w[N], d[M]; cin >> n >> m; for (int i = 1; i <= n; i++) { cin >> w[i] >> d[i]; } for (int i = 1; i <= n; i++) { for (int j = m; j >= w[i]; j--) { dp[j] = max(dp[j-w[i]] + d[i], dp[j]); } } cout << dp[m] << endl; return 0; } |
