一、热身 [Cloned] H - 请使用sort

最新推荐文章于 2024-10-21 09:14:02 发布

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本文介绍了一种算法，用于从多种T恤设计方案中选择最能满足大众口味的组合。通过收集人们对不同设计元素的满意度评分，算法能找出得分最高的设计方案集合。

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原题：

Soon after he decided to design a T-shirt for our Algorithm Board on Free-City BBS, XKA found that he was trapped by all kinds of suggestions from everyone on the board. It is indeed a mission-impossible to have everybody perfectly satisfied. So he took a poll to collect people's opinions. Here are what he obtained: N people voted for M design elements (such as the ACM-ICPC logo, big names in computer science, well-known graphs, etc.). Everyone assigned each element a number of satisfaction. However, XKA can only put K (<=M) elements into his design. He needs you to pick for him the K elements such that the total number of satisfaction is maximized.

题意：

假设用n种T恤，然后m个人分别对这n种T恤打分，要计算出哪几种T恤的得分和最高。

题解：

创建结构体包括：T恤序号，T恤得分总和

输入数据依次得到T恤的得分总和，排序输出就得了

（这题解真的太糊弄了，主要也是前边的这些题就是模拟一下排序一下就好了）

主要学习一下结构体的sort的用法

就酱

代码：AC

#include<iostream>
using namespace std;
struct nom
{
	int No=0;
	double sat=0;
};
int n, m, k;
double **bank;
nom *ans;
 
int main(void)
{
	while (cin >> n >> m >> k)
	{
		bank = new double*[n];
		ans = new nom[m];
		for (int i = 0; i < n; i++)
		{
			bank[i] = new double[m];
			for (int j = 0; j < m; j++)
			{
				cin >> bank[i][j];
				ans[j].No = j+1;
				ans[j].sat += bank[i][j];
			}
		}
		for (int i = 1; i < m; i++)
		for (int j = 0; j < m - i;j++)
		{
			if (ans[j].sat < ans[j + 1].sat)
			{
				nom s = ans[j]; ans[j] = ans[j + 1]; ans[j + 1] = s;
			}
		}
		for (int i = 1; i < k; i++)
		for (int j = 0; j < k - i; j++)
		{
			if (ans[j].No < ans[j + 1].No)
			{
				nom s = ans[j]; ans[j] = ans[j + 1]; ans[j + 1] = s;
			}
		}
		for (int i = 0; i < k; i++)
		{
			cout << ans[i].No ;
			if (i < k - 1)
				cout << ' ';
			else
				cout << endl;
		}
	}
	return 0;
}