0801笔试

求最大的重叠区间个数

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
class Node
{
public:
	int value;
	int flag;
	Node(int value, int flag) :value(value), flag(flag) {}
	bool operator<(const Node&node)
	{
		if (this->value == node.value)
			return this->flag > node.flag;

		return this->value < node.value;

	}

};

int main()
{
	int N;
	cin >> N;
	vector<Node>myvec;


	for (int i = 0; i < N; i++)
	{
		int begin, end;
		cin >> begin;
		myvec.push_back(Node(begin, 0));
		cin >> end;
		myvec.push_back(Node(end,1));
	}
	sort(myvec.begin(),myvec.end());
	int overlap_num = 0;
	int res = 0;
	for (auto&elem:myvec)
	{
		if (elem.flag == 0)
		{
			overlap_num++;
			res = max(overlap_num, res);
		}
		else
		{
			overlap_num--;
		}
	}
	cout << res << endl;
	system("pause");
	return 0;
}

简单的完全背包问题

#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;

int main()
{
	int total_money;
	cin >> total_money;
	int n;
	cin >> n;

	vector<int>prices(n + 1, 0);
	vector<int>values(n + 1, 0);
	for (int i = 1; i <= n; i++)
	{
		cin >> prices[i];
		cin >> values[i];
	}
	
	//dp[i][j]表示当预算为j时，前i件物资的最优组合对应的最大价值
	vector<vector<int>>dp(n + 1, vector<int>(total_money + 1, 0));
	for(int i=1;i<=n;i++)
		for (int j = 1; j <= total_money; j++)
		{
			if (j < prices[i])
			{
				dp[i][j] = dp[i - 1][j];
			}
			else
			{
				dp[i][j] = max(dp[i][j - prices[i]] + values[i],max(dp[i - 1][j], dp[i-1][j - prices[i]] + values[i]));
			}
		}

	cout << dp[n][total_money] << endl;
	system("pause");
	return 0;
}