3 最优调度问题

作者: 赵晓鹏时间限制: 1S章节: 动态规划与贪心

 

 

输入范例 :

2
1 4
3 1
3
1 6
3 1
7 1
3
1 6
3 2
7 1
 

输出范例 :

10
21
23
 

#include
#include
#include
#include
using namespace std;

#define MAXMUM 50000;

long long tsum = 0;
struct student {
    long long start;
    long long process;
};
bool comparable1(student a, student b) {
    return a.start < b.start;
}


int main()
{
  
    long long result = 0;
    long long ptr = 0;
    long long excuTime = 0;  //执行时间
    long long interval = 0;  //当前开始时间和下一个任务的开始时间的间隔
    
    long long length;
    student arr[50000];
    priority_queue, greater> q;
    for (; cin >> length;) {
        long long answer = 0;
        for (long long i = 0; i < length; i++) 
        {
            cin >> arr[i].start;
            cin >> arr[i].process;

        }
        sort(arr, arr + length, comparable1);
        long long  temperate = arr[0].start;
        for (long long i = 1; i < length; i++) 
        {
            q.push(arr[i - 1].process);
            if (temperate != arr[i].start) 
            {
                interval = arr[i].start - temperate;
                 ptr = temperate;
                while (!q.empty())
                {
                    if (interval > 0) {
                        if (q.top() > interval) {
                            long long s = q.top();
                            q.pop();
                            s -= interval;
                            q.push(s);
                            break;

                        }
                        else {
                            interval -= q.top();
                            answer += q.top() + ptr;
                            ptr += q.top();
                            q.pop();
                        }
                    }
                    else 
                    {
                        break;
                    }
                }
                temperate = arr[i].start;
            }
        }
        q.push(arr[length - 1].process);
        while (1) 
        {
            if (!q.empty()) {
                answer += q.top() + temperate;
                temperate += q.top();
                q.pop();
            }
            else {
                break;
            }
        }
        cout << answer << endl;

    }

    return 0;
}