poj3624 Charm Bracelet

Charm Bracelet

Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 44500 Accepted: 19078
Description

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

题目解析：

这个题目是一个很简单的dp问题，也是dp中最常见的01背包模板题目，首先灌输一下什么是01背包问题，01背包问题就是说有n个物品，每个物品数量是有限制的，每个物品有一个价值，给你一个容量为W的背包，问可以装物品的最大价值是多少，对于这个题目也是一样，首先我们可以设dp[i]表示容量为i的背包最多可以装物品的价值，所以很容易得到动态转移方程，dp[i] = max(dp[i],dp[i-w[j]]);
有什么不懂得可以直接评论或者私聊我。
下面将会给出c++和java两种代码，菜鸟代码，大佬勿喷，有错误请指出，谢谢合作

c++代码：

#include 
#include 
#include 
using namespace std;
const int maxn = 200000;
int w[maxn],v[maxn],dp[maxn];
int main()
{
    int n,m;
    while(cin>>n>>m)
    {
        for(int i=0; icin>>w[i]>>v[i];
        int ans = 0;
        for(int i=0; ifor(int j=m; j>=w[i]; j--)
            {
                dp[j] = max(dp[j],v[i]+dp[j-w[i]]);
                ans = max(ans,dp[j]);
            }
        cout<return 0;
}

java代码

import java.util.Scanner;

public class Main {
    public static void main(String args[]){
        int n = 0,W = 0,ans = 0;
        Scanner cin = new Scanner(System.in);
        int []v = new int[200000];
        int []w = new int[200000];
        int []dp = new int[200000];
        n = cin.nextInt();
        W = cin.nextInt();
        for(int i=0;ifor(int i=0;ifor(int j=W;j>=w[i];j--){
                dp[j] = Math.max(dp[j],dp[j-w[i]]+v[i]);
                ans = Math.max(ans,dp[j]);
            }
        System.out.println(ans);
    }
}