背包问题

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 iin 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.

输入

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

输出

Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints

样例输入

4 61 42 63 122 7

样例输出

23

典型的背包问题，第一次在主函数里数组开太大，改成全局数组

#include<iostream>#include<cstring>using namespace std;int D[3405],W[3405];int A[12885];int main(){   int N,M,j,i;   cin >> N >> M;   for(i = 1; i <= N; ++i)   cin >> W[i] >> D[i];   memset(A,0,sizeof(A));   for(i = 1; i <= N; ++i)   {   for(j = M; j >= W[i]; --j)   {   if(A[j-W[i]] + D[i] > A[j])   A[j] =  A[j-W[i]] + D[i];   else   A[j] = A[j];   }   }   cout << A[M];        system("pause");   return 0;}