百炼：4131:Charm Bracelet

4131: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 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

  • 23
    

  • \

  • 非常明显的01背包问题，使用动态规划方式解决。发现这个题非常的蛋疼，使用vector就会超时，使用数组就accepted，这个感觉不够科学。具体的介绍可以看我的另外一篇博文：

  • **blog.csdn.net/qq_24369113…
    **

  • 代码如下：

  • #include<iostream>
    #include<vector>
    #include<string>
    #include<map>
    #include<queue>
    
    using namespace std;
    
    int main()
    {
    	int n, m;
    	cin >> n >> m;
    	int d[2][12881];
    	int things[3405][2];
    	//vector<vector<int>> d(n, vector<int>(m+1, 0));
    
    //	vector<vector<int>> things(n, vector<int>(2, 0));
    	for (int i = 0; i < n; i++)
    	{
    		int tw, td;
    		cin >> tw >> td;
    		things[i][0] = tw;
    		things[i][1] = td;
    
    	}
    
    	for (int i = 0; i < m+1; i++)
    	{
    		if (things[0][0]>i)
    		{
    			d[0][i] = 0;
    		}
    		else
    			d[0][i] = things[0][1];
    	}
    
    	for (int i = 1; i < n; i++)
    	{
    		for (int j = 0; j < m + 1; j++)
    		{
    			int tw = things[i][0];
    			int td = things[i][1];
    			
    			if (j >= tw)
    				d[i%2][j] = max(d[abs(i % 2 - 1)][j - tw] + td, d[abs(i % 2 - 1)][j]);
    			else
    				d[i%2][j] = d[abs(i % 2 - 1)][j];
    		}
    	}
    
    	cout << d[abs(n%2-1)][m] << endl;
    	system("pause");
    }
    

  • \