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

简单模拟题，注意读懂题意就行

#include <iostream>#include <queue>using namespace std;#define CUSTOMER_MAX 1000+1#define INF 0x6fffffff  #ifndef LOCAL//#define LOCAL#endif LOCALint n; // number of windows <=20int m ;// queue capacity  <=10int k; // customers  <=1000int q; // query times <=1000int  ProcessTime[CUSTOMER_MAX]; // queue<int> que[20];queue<int >Wait[20];int currTime = 0;int LeaveTime[CUSTOMER_MAX];int Timebase[20] = {0};int main(){#ifdef LOCALfreopen("input.txt","r",stdin);freopen("output.txt","w",stdout);#endifcin>>n>>m>>k>>q;for(int i=0;i<k;i++){cin>>ProcessTime[i];}int index;int top = 0;for(int i = 0;i<2*k;i++) {int min_len  = m;if(top !=k) // if there are any customer not in line{for(int j=0;j<n;j++) {if(min_len > que[j].size() ){min_len = que[j].size();index = j;}}}if(min_len != m) // find minimum queue {que[index].push(top);Wait[index].push(ProcessTime[top]);top++;}else  // no queue available or no customer not in line, then customer pop {long min_wait = INF;bool empty = true;for(int j=0;j<n;j++){if(Wait[j].empty()) continue;if(min_wait > Timebase[j]+Wait[j].front())  // find current minimum wait time{min_wait = Timebase[j]+Wait[j].front();index = j;empty = false;}}if(empty) break;Timebase[index] += Wait[index].front();LeaveTime[que[index].front()] = Timebase[index];que[index].pop();Wait[index].pop();}}//60*9int qq;for(int i=0;i<q;i++){cin>>qq;qq--;if(LeaveTime[qq]-ProcessTime[qq]<60*9){int hour = LeaveTime[qq]/60;int second = LeaveTime[qq]%60;printf("%02d:%02d\n",8+hour,second);  }elseprintf("Sorry\n");}#ifdef  LOCALsystem("PASUE");#endif LOCALreturn 0;}