就地滚动处理0-1背包

I - 01背包（就地滚动）

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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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 weight W_i (1 ≤ W_i ≤ 400), a 'desirability' factor D_i (1 ≤ D_i ≤ 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: W_i and D_i

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

#include<iostream>using namespace std;int f[12881];  //f[k]表示：当背包的容量为k时的最大价值int max(int a,int b){return a>b?a:b;}int main(){int Num,TotalWeight,i,k,j,weight[12881],value[12881];        cin>>Num>>TotalWeight;f[0]=0;for(i=0;i<Num;i++)cin>>weight[i]>>value[i];for(i=0;i<Num;i++)for(k=TotalWeight;k>=weight[i];k--){f[k]=max(f[k],(f[k-weight[i]]+value[i]));}cout<<f[TotalWeight]<<endl;return 0;}