POJ #1025 Department

模拟题。

这题第一个障碍是现在少见的循环电梯 （'pater-noster' elevator)

"The building has `pater-noster' elevator, i.e. elevator build up from several cabins running all around."

这种叫做 paster-noster 的电梯是 running all around 的，如动画所示，题目中费解的地方便清楚了。（欲知详情，请往维基百科）

第二个难点：模拟

模拟的核心是排队——等电梯或等房间

一开始可能感觉无从下手，模拟题的分析方法最主要的是分析事件。

准确地说，这道题有且仅有两个事件

排队等候（wating in queue)

行动（progress)

题目要求输出的是特工的活动记录（record）

solution:

1.用结构体表示事件：排队等候E，行动S

2.将电梯也看做房间编号是XX00

3.用优先队列维护排队等候的序列priority_queue<E, vector<E> > que,为此还需定义小于(<)运算符

bool operator < (const E& a,  const E& b){ ...}

4.维护vector<S> record[26]：特工的行动序列

5.房间状态的维护

（1）room[11][11] ：房间空闲的起始时刻，初始化为0，模拟过程中要不断更新

（2）updated[11][11]: updated[i][j]表示room[i][j]是否更新过

若当前等待事件的目标房间（或电梯）更新过，则要将当前等待事件再次入队

6.处理过程中时间统一化成秒，输出时再转化成指定的格式

7.其他细节见代码

网上见到的题解大多代码过长，可读性差，我自己写了个比较简明的版本，200行多一点

#include<bits/stdc++.h>
using namespace std;

typedef pair<int,int> P;

int room[11][11]; //room[i][j]:房间空闲的起始时刻
bool updated[11][11]; //起始时刻是否已更新

vector<P> agent[26];

int time[26], sta[26];

int done[26];//已经访问的房间数目

struct S //行程
{
    int des;
    int cost;
    S(int des, int cost):des(des), cost(cost){}
};

struct E   //排队
{
    int code;
    int beg;
    int pos;
    E(int code, int beg, int pos):code(code), beg(beg), pos(pos){};
};

bool operator <(const E &a, const E &b)
{
    //两人不是站在同一队列中等待
    if(a.pos!=b.pos) return a.beg > b.beg;
    //两人同时到达
    if(a.beg==b.beg) return a.code > b.code;

    int f=a.pos/100, r=a.pos%100;

    //等电梯
    if(a.pos%100==0)
    {
        int t1=(a.beg%5?(a.beg/5+1)*5:a.beg);
        int t2=(b.beg%5?(b.beg/5+1)*5:b.beg);
        //两人乘同一班电梯，资历高的先进
        if((t1<=room[f][r]&&t2<=room[f][r])||t1==t2)
            return a.code>b.code;
        return t1>t2;
    }
    else //等房间
    {
        //两人都得等，资历高的先进
        if(a.beg<=room[f][r]&&b.beg<=room[f][r]) return a.code>b.code; //!room[f][r]更新会影响这种比较
        //至少有一个人不用等，先到的先进
        else return a.beg > b.beg;
    }
}

priority_queue<E, vector<E> >que; //排队
vector<S> record[26]; //行程

void input()
{
    FILE* fp=stdin; //提交时改成stdin
    char c;
    int h, m, s, pos, dur;
    while(fscanf(fp," %[A-Z] ",&c))
    {
        int idx=c-'A';
        fscanf(fp,"%d:%d:%d",&h,&m,&s);
        time[idx]=sta[idx]=3600*h+60*m+s;

        while(fscanf(fp,"%d",&pos),pos)
        {
           fscanf(fp,"%d",&dur);
           agent[idx].push_back(P(pos, dur));
        }
        agent[idx].push_back(P(pos, 0)); //! Exit
    }
}

void init()
{
    for(int i=0; i<26; i++)
    {
        if(!agent[i].size()) continue;
        if(agent[i][0].first/100==1)
            record[i].push_back(S(agent[i][0].first,30));//往一层某房间
        else record[i].push_back(S(100,30));//往一层电梯
        time[i]+=30;
        que.push(E(i,time[i],record[i].back().des));
    }
}

void simulator()
{
    int wait, id, cost, to, f, r;
    while(!que.empty())
    {
        E e=que.top();
        que.pop();
        id=e.code;
        f=e.pos/100;
        r=e.pos%100;
        if(updated[f][r])
        {
            que.push(e);
            updated[f][r]=false;
            continue;
        }

        if(r==0) //等电梯
        {
            if(e.beg<=room[f][r]) wait=room[f][r]-e.beg;
            else wait=(e.beg%5? 5-e.beg%5: 0);
            if(wait) record[id].push_back(S(e.pos, wait));
            time[id]+=wait;
            room[f][r]=time[id]+5;
            updated[f][r]=true;
            to=agent[id][done[id]].first;
            if(to==0) //!Exit
            {
                cost=30*(e.pos/100-1);
                record[id].push_back(S(100, cost));
                record[id].push_back(S(to,30));
                continue;
            }
            cost=30*(max(to/100,e.pos/100)-min(to/100,e.pos/100));
            record[id].push_back(S(to/100*100, cost)); //在电梯里
            time[id]+=cost;
            record[id].push_back(S(to, 10)); //往房间
            time[id]+=10;
            que.push(E(id,time[id],to));
        }
        else //等房间
        {
            wait=max(room[f][r]-e.beg,0);
            if(wait) record[id].push_back(S(e.pos,wait));
            time[id]+=wait;
            cost=agent[id][done[id]].second;
            record[id].push_back(S(-e.pos, cost)); //取反
            time[id]+=cost;

            room[f][r]=time[id];
            updated[f][r]=true;

            done[id]++;
            to=agent[id][done[id]].first;
            if(to==0&&f==1) //!Exit
            {
                record[id].push_back(S(to, 30));
                continue;
            }
            if(to/100!=f) to=f*100; //需等电梯
            record[id].push_back(S(to, 10)); //往电梯
            time[id]+=10;
            que.push(E(id,time[id],to));
        }
    }
}

void t_p(int t)
{
    int h=t/3600, m=(t%3600)/60, s=t%3600%60;
    printf("%02d:%02d:%02d ",h,m,s);
}

void e_p(int pre, int cur)
{
    if(pre==0) {printf("Entry\n"); return;}
    if(cur==0) {printf("Exit\n"); return;}
    if(cur<0) {printf("Stay in room %04d\n",-cur); return;}
    if(pre==cur)
    {
        if(cur%100==0) printf("Waiting in elevator queue\n");
        else printf("Waiting in front of room %04d\n",cur);
        return;
    }
    if(pre%100==0&&cur%100==0) {printf("Stay in elevator\n"); return;};
    if(pre<0)
    {
        if(cur%100==0) printf("Transfer from room %04d to elevator\n",-pre);
        else printf("Transfer from room %04d to room %04d\n",-pre, cur);
        return;
    }
    else printf("Transfer from elevator to room %04d\n",cur);
}

void output()
{
    int pre_pos, cur_pos, t;
    for(int i=0; i<26; i++)
    {
       if(!record[i].size()) continue;
       t=sta[i];
       pre_pos=0;

       printf("%c\n",'A'+i);
       for(int j=0; j!=record[i].size(); j++)
       {
           t_p(t);
           t_p(t+=record[i][j].cost);
           cur_pos=record[i][j].des;
           e_p(pre_pos, cur_pos);
           pre_pos=cur_pos;
       }
       printf("\n");
    }
}

int main()
{
    input();
    init();
    simulator();
    output();
    return 0;
}