习题8-12(uva-1153)

最新推荐文章于 2021-07-01 11:56:40 发布

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本文介绍了一个使用优先级队列处理任务调度的问题。通过读取任务的完成时间和优先级，对任务进行排序并分配资源。算法考虑了时间窗口和任务更新，展示了如何在C++中实现这一过程。

#include <iostream>
#include <istream>
#include <sstream>
#include <vector>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <cstring>
#include <unordered_map>
#include <unordered_set>
#include <algorithm>
#include <numeric>
#include <chrono>
#include <ctime>
#include <cmath>
#include <cctype>
#include <string>
#include <cstdio>
#include <iomanip>


#include <thread>
#include <mutex>
#include <condition_variable>
#include <functional>
#include <iterator>
using namespace std;
const int maxn = 8e5 + 7;
struct Task
{
	int d,q;
	bool operator<(const Task& input) {
		if (d == input.d) return q < input.q;
		return d < input.d;
	}
}task[maxn];
int n;

priority_queue<int, vector<int>, less<int> > big;

int main()
{
	int t,a,b,kCase = 0;
	cin >> t;
	while (t--) {
		cin >> n;
		for (int i = 0; i < n; i++) {
			cin >> task[i].q >> task[i].d;
		}
		sort(task, task + n);

		int cur = 0,ret = 0;
		while (!big.empty()) big.pop();

		for (int i = 0; i < n; i++) {
			if (cur + task[i].q <= task[i].d) {
				cur += task[i].q;
				big.push(task[i].q);
				ret++;
			}
			else if(!big.empty()){
				int top = big.top();
				if (top > task[i].q) {
					cur += task[i].q - top;
					big.pop();
					big.push(task[i].q);
				}
			}
		}
		if (kCase++) cout << endl;
		cout << ret << endl;
	}
	return 0;
}