01背包（就地滚动）

<span style="color:#3333ff;">/*__________________________________________________________________________________________________*     copyright:   Grant Yuan                                                                     **     algorithm:   01背包（就地滚动）                                                             **     time     :   2014.7.18                                                                      **     declare  :   题目中说N最大是3400多，但是一开始开了5000内存还是运行时错误，后来直接改了50000 *                                                       *                                                                                                 **_________________________________________________________________________________________________*</span>

<span style="color:#3333ff;">I - 01背包（就地滚动）Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64uSubmit StatusDescriptionBessie 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 DiOutput* Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraintsSample Input4 61 42 63 122 7Sample Output23*/#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>using namespace std;int w[50001];int p[50001];int sum;int n;int dp[50001];int main(){cin>>n>>sum;for(int i=0;i<n;i++)cin>>w[i]>>p[i];memset(dp,0,sizeof(dp));for(int i=0;i<n;i++)   for(int j=sum;j>=w[i];j--)    {dp[j]=max(dp[j],dp[j-w[i]]+p[i]);     }cout<<dp[sum]<<endl;return 0;}</span>