** 動態規劃(2) 編程題＃2： Charm Bracelet（Coursera 程序設計與算法 專項課程4；#define maxN 3402 此處沒有 ; 號！）

編程題＃2： Charm Bracelet

來源: POJ (http://bailian.openjudge.cn/practice/4131/)

注意： 總時間限制: 1000ms 內存限制: 65536kB

描述
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 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.

輸入
Line 1: Two space-separated integers: N and M
Lines 2..N+1: Line i+1 describes charm i with two space-separated integers: Wi and Di

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

樣例輸入

4 6
1 4
2 6
3 12
2 7

樣例輸出

23

解題思路：
dd大牛的揹包九講（第一講 01揹包問題）http://blog.csdn.net/happyygdx/article/details/79126439
http://blog.csdn.net/iamiman/article/details/53698222

程序解答：

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;

#define maxN 3402   //注：此處沒有 ; 號！
#define maxM 12880

int main(){
    int N, M, W[maxN + 1], D[maxN + 1];
    int f[maxM + 1];
    cin >> N >> M;
    for (int i = 1; i <= N; i++)
        cin >> W[i] >> D[i];
    memset(f, 0, sizeof(f));

    for (int i = 1; i <= N; i++){
        for (int j = M; j >= W[i]; j--)
            f[j] = max(f[j], f[j - W[i]] + D[i]);
    }

    cout << f[M] << endl;

    return 0;
}