POJ 3624 Charm Bracelet （01背包）

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

大概思路：

本来是想通过一个二维数组进行记录，但因为本题的要求，用二维数组会超内存。所以利用滚动数组进行记录以节省空间。 这种01背包类的问题的状态转移方程为（i为第i个物体，j为剩下可用的容量）： dp[i][j]=Max(dp[i-1][j],dp[i-1][j-weight[i]]+value[i]) 即第i个物品的取与不取，这个判断的前提是j>=weight[i]，否则dp[i][j]=dp[i-1][j]。 而利用滚动组节省空间的方法则是从有开始往左边扫，这样就不会出现错误的情况。

代码：

#include<iostream> #include<algorithm> #include<cstdio> using namespace std; int dp[13000]; int main() { // freopen("in.txt","r",stdin); int n,m; int w[3500],v[3500]; cin>>n>>m; for(int i=1;i<=n;++i) cin>>w[i]>>v[i]; for(int i=1;i<=m;++i){//初始化边界条件 if(w[1]<=i) dp[i]=v[1]; else dp[i]=0; } for(int i=2;i<=n;++i){ for(int j=m;j>=0;--j){ if(w[i]<=j) dp[j]=max(dp[j],dp[j-w[i]]+v[i]); else dp[j]=dp[j]; } } cout<<dp[m]<<endl; return 0; }