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 题意：最简单的01背包问题 分析：用二维的在前i个中选择重量不大于j的思路会超内存。。。 收获：学习dp第一发，为男人八题做准备。。。 #include <iostream> #include <algorithm> #include <string.h> using namespace std; int n,m; int w[3500],v[3510]; //int dp[50][50]; int dp[15000]; int main () { cin >> m >> n; for(int i=0; i<m; i++) { cin >> w[i] >> v[i]; } memset(dp,0,sizeof(dp)); for(int i=0;i<m;i++) { for(int j=n;j>=w[i];j--) { dp[j]=max(dp[j],dp[j-w[i]]+v[i]); } } cout<< dp[n]; return 0; }