UVA_10154_Weights and Measures

最新推荐文章于 2018-12-05 22:42:41 发布

转载最新推荐文章于 2018-12-05 22:42:41 发布 · 218 阅读

文章标签：

本文通过一个具体的 C++ 实现案例介绍了如何使用动态规划解决特定问题。该示例涉及 C++ 中多种数据结构的应用，如向量(vector)、配对(pair)等，并通过自定义类和操作符来组织数据。程序主要关注如何通过动态规划方法找到最优解。

#include<iostream>
#include<sstream>
#include<string>
#include<vector>
#include<list>
#include<set>
#include<map>
#include<stack>
#include<queue>
#include<algorithm>
#pragma warning(disable:4996)
using std::cin;
using std::cout;
using std::endl;
using std::stringstream;
using std::string;
using std::vector;
using std::list;
using std::pair;
using std::set;
using std::multiset;
using std::map;
using std::multimap;
using std::stack;
using std::queue;
using std::priority_queue;
class Turtle 
{
public:
	int weight, strength;
	friend bool operator < (const Turtle &a, const Turtle &b) 
	{
		return a.strength < b.strength;
	}
	Turtle()
	{
		weight = 0;
		strength = 0;
	}
	Turtle(const int &w, const int &s)
	{
		weight = w;
		strength = s;
	}
};
int main()
{
	//freopen("input.txt", "r", stdin);  
	//freopen("output.txt", "w", stdout);  
	vector<Turtle>turtle(1);
	int weight, strength;
	while (cin >> weight >> strength)
	{
		turtle.push_back({ weight,strength });
	}
	std::sort(turtle.begin()+1, turtle.end());
	vector<vector<int>>dp(turtle.size(), (vector<int>)turtle.size());
	for (size_t i = 0; i < turtle.size(); i++)
	{
		dp[i][0] = 0;
		for (int j = 1; j < turtle.size(); ++j)
		{
			dp[i][j] = 2000000000;
		}
	}
	for (size_t i = 1; i <turtle.size(); i++)
	{
		for (size_t j = 1; j <= i; j++)
		{
			dp[i][j] = dp[i - 1][j];
			if (turtle[i].strength - turtle[i].weight >= dp[i - 1][j - 1] && dp[i - 1][j - 1] != 2000000000)
			{
				dp[i][j] = std::min(dp[i][j], dp[i - 1][j - 1] + turtle[i].weight);
			}
		}
	}
	for (size_t i = turtle.size()-1; i >= 1; i--)
	{
		if (dp[turtle.size() - 1][i] != 2000000000)
		{
			cout << i << endl;
			break;
		}
	}
	return 0;
}