dp Charm Bracelet

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 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.

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: W_i and D_i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

我的Runtime error代碼

#include<iostream>
#include<cstring>
#include<vector>
#include<algorithm>
using namespace std;
struct Node{
    int w;
    int d;
    double dw;
    Node(){}

}node[3410];
bool cp(Node a,Node b){
        if(a.dw>b.dw)
            return true;
        else
            if(a.dw==b.dw&&a.w<b.w)
                return  true;
}
int main(){
    int n,m;
    cin>>n>>m;
    for(int i=0;i<n;i++){
        cin>>node[i].w;
        cin>>node[i].d;
        node[i].dw=1.0*node[i].d/node[i].w;
    }
    sort(node,node+n,cp);
    int ct=0;
    for(int i=0;i<n;i++){
        if(node[i].w<=m){
            ct+=node[i].d;
            m-=node[i].w;
        }else continue;
    }
    cout<<ct;
}

//意識到是dp了，但是還是不會寫。

大佬的AC代碼

// 揹包問題（動態規劃）
#include <iostream> 
#include <cstdio>
#include <cstring>
#define MAXN 3402
#define MAXM 12880
using namespace std;

int main(){
    int N, M, W[MAXN+5], D[MAXN+5], dp[MAXM+5];
    while(scanf("%d%d", &N, &M) != EOF){
        for(int i=0; i<N; i++){
            scanf("%d%d", &W[i], &D[i]);

        }
        memset(dp, 0, sizeof(dp));
        for(int i=0; i<N; i++){
            for(int left_w=M; left_w>=W[i]; left_w--){
                dp[left_w] = max(dp[left_w-W[i]]+D[i], dp[left_w]);
            }
        }
        printf("%d\n", dp[M]);
    }   
    return 0;
}

PS：dp數組記錄當容量爲當前下標時，最高的價值；dp數組是動態覆蓋的；每放一個物品就會更新一遍dp數組；當放第i個物品的時候，i-1及其之前的都已經計算好了。