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 (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 7Sample Output
23 #include<iostream>using namespace std;const int maxn = 100 * 3500;int f[maxn];int M,N;int c[3403];int w[3403];int max(int a,int b){if(a>b)return a;return b;}void ZeroOnePack(int cost,int weight){ for (int i = M; i>= cost; i--) f[i] = max(f[i], f[i - cost] + weight);}int main(){ int i; cin >> N >> M; memset(f, 0 ,sizeof(f)); for (i = 1; i<= N; i++) cin >> c[i] >> w[i]; for (i = 1; i <= N; i++) ZeroOnePack(c[i],w[i]); cout << f[M] << endl; return 0;}转载于:https://www.cnblogs.com/w0w0/archive/2011/11/23/2259848.html
相关资源：数据结构—成绩单生成器