PAT (Advanced Level) 1014. Waiting in Line (30)

简单模拟题。

#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<cstdio>
#include<map>
#include<queue>
using namespace std;

struct X
{
    int st;
    int len;
    int en;
}p[1500];
queue<int>Q[25];
int n,m,k,s;

bool check()
{
    for(int i=1;i<=n;i++)
        if(!Q[i].empty()) return 1;
    return 0;
}

int main()
{
    scanf("%d%d%d%d",&n,&m,&k,&s);
    for(int i=1;i<=k;i++) scanf("%d",&p[i].len);
    int pos;
    for(pos=1;pos<=min(k,n);pos++) { p[pos].st=480; Q[pos].push(pos); }
    for(;pos<=min(n*m,k);pos++)
    {
        int id=pos%n; if(id==0) id=n;
        Q[id].push(pos);
    }
    while(1)
    {
        if(check()==0) break;
        int Min=999999;
        for(int i=1;i<=n;i++)
        {
            if(Q[i].empty()) continue;
            int id=Q[i].front();
            Min=min(Min,p[id].st+p[id].len);
        }

        for(int i=1;i<=n;i++)
        {
            if(Q[i].empty()) continue;
            int id=Q[i].front();
            if(p[id].st+p[id].len==Min)
            {
                p[id].en=p[id].st+p[id].len;
                Q[i].pop();
                if(pos<=k) Q[i].push(pos++);
                if(!Q[i].empty()) p[Q[i].front()].st=p[id].en;
            }
        }
    }
    for(int i=1;i<=s;i++)
    {
        int id; scanf("%d",&id);
        if(p[id].st/60>=17) printf("Sorry\n");
        else printf("%02d:%02d\n",p[id].en/60,p[id].en%60);
    }
    return 0;
}

 

posted @ 2016-05-16 10:25  Fighting_Heart  阅读(157)  评论(0编辑  收藏  举报