LightOj1303
As we all know that not only children but Ferris-Wheel (FW) is favorite to all ages. That's why the LightOJ Park has made a special FW for the programmers. And unlike other Ferris-Wheels, this FW has m seats and every seat is different and very special. Since its circular, the seats are numbered from 1 to m in anticlockwise order, so seat 2 is adjacent to seat 1 and seat 3, seat 1 is adjacent to seat 2 and seat m.
One day n programmers (identified by 1 to n) came and each of them wanted to ride in the FW. Since every seat is special, everyone wants to taste the ride of every seat. So, they made the following algorithm:
1. They will form a queue from 1 (front) to n (back). As they want to enjoy the ride, they want to sit alone in a seat, because two (or more) may start problem-solving in the ride.
2. The FW moves in clockwise order. Initially the seat 1 is in bottom and after 5 seconds, it starts. And when it starts, in each 5 second interval the next seat comes to bottom. So, the order of the seats in bottom are seat 1 (at time 5), seat 2 (at time 10), ..., seat m (at time 5*m), seat 1 (at time 5*m+5), ... and in this interval, a person sits in that seat or gets out (if he is in the seat) or both (one gets out and another gets in). Assume that these kinds of movements take quite a small time and thus that can be ignored.
3. When a programmer gets out from one seat (he just rode in that seat), then if he has ridden in all the seats, he will leave, otherwise he will join in the back of the queue.
4. When a seat comes into bottom, if all the programmers in the queue have ridden in the seat, nothing happens. Otherwise the first person (from front) in the queue who hasn't ridden in the seat sits in that seat and other programmers keep standing. For example, the current queue is 5, 2, 3, 1 and a seat is in the bottom which has been already ridden by 5 and 2 but 3 hasn't; so, 3 will sit in that seat leaving the queue: 5, 2, 1.
Now you are the (n+1)th programmer and you are unlucky, because you are assigned a task, and the task is to find the minimum time when you are sure that everyone has ridden in all the seats.
Input
Input starts with an integer T (≤ 400), denoting the number of test cases.
Each case starts with a line containing two integers n (1 ≤ n ≤ 20) and m (2 ≤ m ≤ 20).
Output
For each case, print the case number and the time.
Sample Input |
Output for Sample Input |
| 2 2 3 3 2 |
Case 1: 65 Case 2: 40 |
题目大意：
n个程序员打算坐摩天轮的每个座位，只有当座位在底部的时候才可以下来，排在队伍最后，或者队伍中第一个没坐过这个位置的人坐上去。程序员不能重复做同一座位。
摩天轮每次移动到下个位子需要5分钟，问最少多久使所有程序员坐完所有座位。
这道题直接模拟，训练编程能力。
#include<iostream>
#include<vector>
#include<queue>
using namespace std;
struct Man
{
vector<int>vis;
};
struct Seat
{
int num;
int people;
bool Empty;
};
int n, m;
int getTime()
{
queue<Seat>wl;
Man man[21];
vector<int>wait;
for (int i = 1; i <= m; ++i)
{
Seat a;
a.Empty = true;
a.num = i;
a.people = -1;
wl.push(a);
}
for (int i = 1; i <= n; ++i)
{
wait.push_back(i);
}
int USETIME = 0;
for (USETIME = 1;; USETIME += 1)
{
Seat sit = wl.front();
wl.pop();
int s = sit.num;
if (!sit.Empty)
{
wait.push_back(sit.people);
man[sit.people].vis.push_back(sit.num);
}
int j;
for (j = 0; j<wait.size(); ++j)
{
bool si = true;
for (int p = 0; p<man[wait[j]].vis.size(); ++p)
{
if (man[wait[j]].vis[p] == s)
{
si = false;
break;
}
}
if (si)
break;
}
if (j != wait.size())
{
Seat sea;
sea.Empty = false;
sea.num = s;
sea.people = wait[j];
wait.erase(wait.begin() + j);
wl.push(sea);
}
else
{
Seat se;
se.Empty = true;
se.num = s;
wl.push(se);
}
bool leave = true;
for (int i = 1; i <= n; ++i)
{
if (man[i].vis.size() != m)
{
leave = false;
break;
}
}
if (leave)
break;
}
return USETIME;
}
int main()
{
int cou = 0;
int Time;
cin >> Time;
while (Time--)
{
cin >> n >> m;
cout<<"Case "<<++cou<<": "<< 5 * getTime() << endl;
}
}
这道题更加简单的方法是用一个二维数组去模拟人（一维表示人的序号，另一维表示座位的序号），坐过某位置记作1，没坐过记作0。
然后用一个一维数组表示摩天轮， int i=0; for(;;i=(i+1)%m)来表示循环，可以不用queue，既简化了代码难度，又提高了效率。