POJ3624 Charm Bracelet
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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 weightWi (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 61 42 63 122 7
Sample Output
23这道题是裸的01背包,比较简单,核心是01背包的状态转移方程,只要与动态规划相关,下意识就觉得很难,其实,弄明白核心之后,也挺简单的。
思路:01背包~
#include<iostream>#include<cstdio>#include<cstring>#define MAXN 5000using namespace std; int f[3*MAXN]; //状态数组int d[MAXN]; //每个物品的价值int w[MAXN]; //每个物品的重量int N,M; int main(){ scanf("%d%d",&N,&M); memset(f,0,sizeof(f)); memset(d,0,sizeof(d)); memset(w,0,sizeof(w)); for(int i=1;i<=N;i++) { scanf("%d%d",&w[i],&d[i]); } for(int i=1;i<=N;i++) { for(int j=M;j>=w[i];j--) { f[j]=max(f[j],f[j-w[i]]+d[i]);//对于每个物品,取其 不放这个物品的价值或放这个物品后 中的最大值 //,如果放,所能承受的容量减去放进去的物品的重量表示放进去了这个物品 } } printf("%d\n",f[M]); return 0;}
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