动态规划--01背包 poj4131

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.

题意
N种物品和和容积为M的背包
dii种物品体积wi， 价值di 求解将哪些物品放入背包可以史总价值最大 求最大值
转移方程：

F(i, j) = max(F(i-1, j), F(i-1, j-w[i]) + D[i],              i > 1;
F(i, j) =  D[1],                                             i = 1 && w[1] <= j;
F(i, j) = 0,                                                 i = 1 && w[1] > j;

写出代码：

#include
using namespace std;

class Item{
public:
    int w, val;
}a[3043];

int main(){
    int N, M;
    cin >> N >>M;
    int dp[N+1][M+1];
    for(int i = 0; i < N; i++){
        cin >> a[i].w >> a[i].val;
    }
    memset(dp, 0, sizeof(dp));
    for(int i = 1; i <= N; i++){ //N个物品
        for(int j = 1; j <= M; j++){ //M质量
            if(j < a[i-1].w) //放不下了
                dp[i][j] = dp[i-1][j];
            else
                dp[i][j] = max((dp[i-1][j]), (dp[i-1][j-a[i-1].w] + a[i-1].val));
        }
    }

    cout << dp[N][M] << endl;
}

不出意外
MLE
建议使用滚动数组