微软201610 在线笔试 题目3 Registration Day

题目地址：
https://hihocoder.com/problemset/problem/1401
可提交代码

题目3 : Registration Day

时间限制:10000ms
单点时限:1000ms
内存限制:256MB

描述

It’s H University’s Registration Day for new students. There are M offices in H University, numbered from 1 to M. Students need to visit some of them in a certain order to finish their registration procedures. The offices are in different places. So it takes K units of time to move from one office to another.

There is only one teacher in each office. It takes him/her some time to finish one student’s procedure. For different students this time may vary. At the same time the teacher can only serve one student so some students may need to wait outside until the teacher is available. Students who arrived at the office earlier will be served earlier. If multiple students arrived at the same time they will be served in ascending order by student number.

N new students need to finish his/her registration. They are numbered from 1 to N. The ith student’s student number is Si. He will be arrived at H University’s gate at time Ti. He needs to visit Pi offices in sequence which are Oi,1, Oi,2, … Oi,Pi. It takes him Wi,1, Wi,2, … Wi,Pi units of time to finish the procedure in respective offices. It also takes him K units of time to move from the gate to the first office.

For each student can you tell when his registration will be finished?

输入

The first line contains 3 integers, N, M and K. (1 <= N <= 10000, 1 <= M <= 100, 1 <= K <= 1000)

The following N lines each describe a student.

For each line the first three integers are Si, Ti and Pi. Then following Pi pairs of integers: Oi,1, Wi,1, Oi,2, Wi,2, … Oi,Pi, Wi,Pi. (1 <= Si <= 2000000000, 1 <= Ti <= 10000, 1 <= Pi <= M, 1 <= Oi,j <= M, 1 <= Wi,j <= 1000)

输出

For each student output the time when he finished the registration.

样例提示

Student 1600012345 will be arrived at the gate at time 500. He needs to first visit Office #1 and then Office #2. He will be arrived at office #1 at time 600. He will leave office #1 at time 1100. Then He will arrive at office #2 at 1200. At the same time another student arrives at the same office too. His student number is smaller so he will be served first. He leaves Office #2 at 1700. End of registration.

Student 1600054321 will be arrived at the gate at time 1100. He will be arrived at Office #2 at 1200. Another student with smaller student number will be served first so he waits for his turn until 1700. He leaves Office #2 at 2000. End of registration.

样例输入

2 2 100
1600012345 500 2 1 500 2 500
1600054321 1100 1 2 300

样例输出

1700
2000

使用优先队列的方法解决

代码如下：

#include <iostream>#include <queue>#include <vector>using namespace std;struct StudentInfo{      int index;//第几个学生0,1,2,3,4,...    int ID;//学生学号,1600012345    int time;//学生的结束时间    int goWhich;//学生要去的office序号(vector中的序号0,1,2,3,4)    //因为STL里优先队列是大优先，故自身若要比s更优先，需要time更小，ID更小    bool operator < (const StudentInfo &s) const    {          if(time == s.time) return ID > s.ID;        else return time > s.time;      }  };  struct Student{      int ID;    int arriveTime;    int orderNum;    vector<int> orderOffices;      vector<int> orderTimes;};  int main()  {      priority_queue<StudentInfo> pQueue;    int N,M,K;    cin>>N>>M>>K;    vector<Student> students(N);    for (int i=0; i<N; i++)    {        //输入        cin>>students[i].ID;        cin>>students[i].arriveTime;        cin>>students[i].orderNum;        for (int j=0; j<students[i].orderNum; j++)        {            int temp1,temp2;            cin>>temp1>>temp2;            students[i].orderOffices.push_back(temp1-1);//office序号从0开始            students[i].orderTimes.push_back(temp2);        }        //加入队列        StudentInfo si;        si.index = i;        si.ID = students[i].ID;        si.time = students[i].arriveTime+K;//初始时间为到达时间+K        si.goWhich = 0;//初始从第0个开始        pQueue.push(si);    }    vector<int> res(N,0);    vector<int> officeEndTime(M,0);//M个office的结束时间    while(!pQueue.empty())    {          //取最优值        StudentInfo stuInfo = pQueue.top();          Student &stu = students[stuInfo.index];        pQueue.pop();        int curOffice = stu.orderOffices[stuInfo.goWhich];        int procTime = stu.orderTimes[stuInfo.goWhich];        //若当前office的结束时间小于stu的到达时间，则office会有一段时间闲置，stu到达后可直接处理         if (officeEndTime[curOffice] < stuInfo.time)        {            //更新office结束时间为学生到达时间+处理时间            officeEndTime[curOffice] = stuInfo.time + procTime;              //更新学生到达时间为学生到达时间+学生处理时间+K            stuInfo.time += procTime + K;        }        //若当前office的结束时间大于等于stu的到达时间，则stu需要等待一段时间        else        {              //更新office结束时间为office结束时间+学生处理时间            officeEndTime[curOffice] += procTime;              //更新学生到达时间为office结束时间+学生处理时间            stuInfo.time = officeEndTime[curOffice] + K;          }          if (stuInfo.goWhich == stu.orderNum-1)//若是最后一个点        {            res[stuInfo.index] = stuInfo.time - K;        }        else        {            stuInfo.goWhich ++;              pQueue.push(stuInfo);          }    }      for(int i = 0; i < N; i ++)    {          cout<<res[i]<<endl;    }      return 0;  }