Charm Bracelet

今天药忘吃喽~ 2022-07-14 13:58 32阅读 0赞

题目：

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 *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.

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 6
    1 4
    2 6
    3 12
    2 7

样例输出

个人见解：0-1背包问题。一共拥有N件物品，要放入一个背包里面，总重量不能超过M，Wi表示第i件物品的重量，Di表示第i件物品的价值。求如何选择装入背包中的物品，使得装入背包中的物品的总价值最大？

在选择装入背包的物品中，对每种物品i只有两种选择，即装入或者不装入背包。

1、对于重量大于背包容量的物品自然是不能放入背包中的。

2、对于重量小于等于背包容量 的物品来说，有两种情况。一是放入第i个物品，则背包中物品的价值为i-1个物品放入容量为j-Wi背包中的价值加上第i个物品的价值Di；二是不放入第i个物品，背包的价值为i-1个物品放入容量为j的背包中的价值。取二者中价值最大的作为把i个物品放入容量为j的背包中的最优解。

源代码：

#include<iostream>
    #include<algorithm>
    using namespace std;
    const int N = 3402;
    const int M = 12880;
    int maxm[M] = { 0 };
    int main()
    {
    	int n, m;
    	cin >> n >> m;
    	int W[N], D[N];
    
    	for (int i = 1; i <= n; i++)
    		cin >> W[i] >> D[i];
    
    	for (int i = 1; i <= n; i++)
    	{
    		for (int j = m; j >= W[i]; j--)
    		{
    			maxm[j] = max(maxm[j - W[i]] + D[i], maxm[j]);
    		}
    	}
    	cout << maxm[m] << endl;
    }