Codeforces #594 div1 C/div2 E – Queue in the Train
作者:互联网
题目链接:https://codeforces.com/contest/1239/problem/C
题意:火车上有n位乘客,按照1到n编号,编号为i的人会在ti分钟想去打水。水箱只能供一位乘客使用,每位乘客会使用p分钟。当一位乘客想要去打水时,他会先看编号在他前面的乘客是不是都在座位上,如果有人没在座位上,他会坐下继续等待,否则他会去排队打水。当某一时刻有几位乘客同时想要打水时,编号最小的乘客会前去打水,其他人会坐下继续等待,计算每位乘客打完水的时间。
做法:模拟。按照题意模拟即可,具体实现见代码。首先我们需要一个优先队列判断座位上的人谁先去打水,然后我们需要一个优先队列判断坐在座位上等待的人,接着我们需要一个队列模拟正在排队的人,最后我们需要一个数据结构记录空的座位。按照题意模拟即可,具体实现见代码。
参考代码:
#include <iostream>
#include <queue>
#include <set>
using namespace std;
struct passenger
{
int pos, startime;
} psger[100005];
struct cmp_waiting
{
bool operator()(const passenger &x, const passenger &y)
{
return x.pos > y.pos;
}
};
struct cmp_sitting
{
bool operator()(const passenger &x, const passenger &y)
{
if (x.startime == y.startime)
return x.pos > y.pos;
else
return x.startime > y.startime;
}
};
priority_queue<passenger, vector<passenger>, cmp_sitting> sitting;
priority_queue<passenger, vector<passenger>, cmp_waiting> waiting;
queue<passenger> queuing;
set<int> emptyseat;
long long ret[100005];
int main()
{
int n;
long long p, now;
cin >> n >> p;
for (int i = 1; i <= n; ++i)
{
cin >> psger[i].startime;
psger[i].pos = i;
sitting.push(psger[i]);
}
now = sitting.top().startime;
emptyseat.insert(n + 1);
while (!sitting.empty() || !queuing.empty() || !waiting.empty())
{
if (!queuing.empty())
{
passenger fir = queuing.front();
queuing.pop();
now += p;
ret[fir.pos] = now;
while (!sitting.empty() && sitting.top().startime <= now)
{
passenger sec = sitting.top();
if (sec.pos < *emptyseat.begin())
{
emptyseat.insert(sec.pos);
queuing.push(sec);
}
else
waiting.push(sec);
sitting.pop();
}
emptyseat.erase(fir.pos);
}
if (queuing.empty() && waiting.empty())
now = max(now, (long long) sitting.top().startime);
while (!sitting.empty() && sitting.top().startime <= now)
{
waiting.push(sitting.top());
sitting.pop();
}
int id = *emptyseat.begin();
if (!waiting.empty() && waiting.top().pos < id)
{
queuing.push(waiting.top());
emptyseat.insert(waiting.top().pos);
waiting.pop();
}
}
for (int i = 1; i <= n; ++i)
cout << ret[i] << " ";
return 0;
}
标签:passenger,594,sitting,乘客,pos,Codeforces,Queue,startime,empty 来源: https://www.cnblogs.com/mapleaves/p/11808907.html