POJ 3624 Charm Bracelet(01 揹包)
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
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
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
Source
題解:01揹包裸題,主要是優化,見程式碼,dp[i][j]表示的是前i個物品放入容量為j的容器裡能夠產生的最大價值。
大家可以看這個大牛題解,講的很詳細。
#include<iostream> #include<algorithm> #include<cstring> using namespace std; int dp[13100]; int n,m,wei[13100],val[13100]; int main() { while(cin>>n>>m) { memset(dp,0,sizeof(dp)); for(int i=1;i<=n;i++) { cin>>wei[i]>>val[i]; } for(int i=1;i<=n;i++) for(int j=m;j>=wei[i];j--) { dp[j]=max(dp[j],dp[j-wei[i]]+val[i]); } cout<<dp[m]<<endl; } return 0; }