Suppose a bank has N windows open for service. There is a yellow line in front of the windows which devides the waiting area into two parts. The rules for the customers to wait in line are:
- The space inside the yellow line in front of each window is enough to contain a line with M customers. Hence when all the N lines are full, all the customers after (and including) the (NM+1)st one will have to wait in a line behind the yellow line.
- Each customer will choose the shortest line to wait in when crossing the yellow line. If there are two or more lines with the same length, the customer will always choose the window with the smallest number.
- Customer[i] will take T[i] minutes to have his/her transaction processed.
- The first N customers are assumed to be served at 8:00am.
Now given the processing time of each customer, you are supposed to tell the exact time at which a customer has his/her business done.
For example, suppose that a bank has 2 windows and each window may have 2 custmers waiting inside the yellow line. There are 5 customers waiting with transactions taking 1, 2, 6, 4 and 3 minutes, respectively. At 08:00 in the morning, customer1 is served at window1while customer2 is served at window2. Customer3 will wait in front of window1 and customer4 will wait in front of window2. Customer5 will wait behind the yellow line.
At 08:01, customer1 is done and customer5 enters the line in front of window1 since that line seems shorter now. Customer2 will leave at 08:02, customer4 at 08:06, customer3 at 08:07, and finally customer5 at 08:10.
Input
Each input file contains one test case. Each case starts with a line containing 4 positive integers: N (<=20, number of windows), M (<=10, the maximum capacity of each line inside the yellow line), K (<=1000, number of customers), and Q (<=1000, number of customer queries).
The next line contains K positive integers, which are the processing time of the K customers.
The last line contains Q positive integers, which represent the customers who are asking about the time they can have their transactions done. The customers are numbered from 1 to K.
Output
For each of the Q customers, print in one line the time at which his/her transaction is finished, in the format HH:MM where HH is in [08, 17] and MM is in [00, 59]. Note that since the bank is closed everyday after 17:00, for those customers who cannot be served before 17:00, you must output "Sorry" instead.
Sample Input2 2 7 5 1 2 6 4 3 534 2 3 4 5 6 7Sample Output
08:07 08:06 08:10 17:00
Sorry
題意：銀行有winNum個窗口，每個窗口的黃線內可以容納maxCap個人。顧客每次都是挑選黃線內人數最少的窗口進入。
分析：模擬隊列。
#include <iostream>
#include <vector>
#include <string>
#include <queue>
#include <algorithm>
using namespace std;
int maxCap;
struct win{
int id; //窗口id
int num; //該窗口目前黃線內的人數
int firstFinish; //該窗口排在第一個的顧客業務結束時的時間
queue<int> cus; //該窗口受理的顧客
};
bool cmp1(win a, win b){
if(a.num != b.num){
if(a.num<b.num)
return true;
else
return false;
}else if(a.num==b.num && a.id!=b.id){
if(a.id<b.id)
return true;
else
return false;
}
}
bool cmp2(win a, win b){
if(a.firstFinish != b.firstFinish){
if(a.firstFinish < b.firstFinish)
return true;
else
return false;
}else if(a.firstFinish == b.firstFinish && a.id != b.id){
if(a.id < b.id)
return true;
else
return false;
}
}
struct customer{
int proTime; //業務處理時間
int finishTime; //最終處理好的時間
};
int main(int argc, char** argv) {
int winNum, cusNum, queNum;
scanf("%d%d%d%d",&winNum, &maxCap, &cusNum, &queNum);
int i;
vector<win> wins;
for(i=0; i< winNum; i++){
win w;
w.id = i;
w.num = 0;
w.firstFinish = 0;
wins.push_back(w);
}
vector<customer> cus(cusNum+1);
int time;
for(i=1; i<=cusNum; i++){
scanf("%d",&time);
cus[i].proTime = time;
sort(wins.begin(),wins.end(),cmp1);
if(wins[0].num < maxCap){
if(wins[0].firstFinish == 0){
wins[0].firstFinish += time;
}
wins[0].num ++;
wins[0].cus.push(i);
}else{
sort(wins.begin(), wins.end(), cmp2);
int index = wins[0].cus.front();//這個窗口第一個顧客
cus[index].finishTime = wins[0].firstFinish;
wins[0].cus.pop();
wins[0].cus.push(i);
index = wins[0].cus.front();
wins[0].firstFinish += cus[index].proTime;
wins[0].num++;
}
}
for(i=0; i<winNum; i++){
win w = wins[i];
while(!w.cus.empty()){
int index = w.cus.front();
cus[index].finishTime = w.firstFinish;
w.cus.pop();
if(!w.cus.empty())
w.firstFinish = w.firstFinish + cus[w.cus.front()].proTime;
}
}
int index;
for(i=0; i<queNum; i++){
scanf("%d",&index);
int time = cus[index].finishTime;
int proTime = cus[index].proTime;
if((time-proTime)>=(17-8)*60){
//如果受理時間是17：00之後的，則輸出sorry
printf("Sorry\n");
}else{
int h = 8 + time/60;
int m = time%60;
printf("%02d:%02d\n",h,m);
}
}
return 0;
}
以下是另一種寫法，但是有兩組數據過不了，待改進：
#include <iostream>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;
int N,M,K,Q;
struct customer{
customer(){
h=8; m=0;
}
int h,m,pro;
};
vector<customer> cus;
struct windows{
windows(){
h=8; m=0;
}
int id;
int h,m;
queue<int> c;
bool operator > (const windows &b)const{
if(c.size()!=b.c.size()){
return c.size() > b.c.size();
}else if((h*60+m)!=(b.h*60+b.m)){
return (h*60+m)>(b.h*60+b.m);
}else{
return id > b.id;
}
}
};
priority_queue<windows, vector<windows>, greater<windows> > win;
int main(){
scanf("%d%d%d%d",&N,&M,&K,&Q);
cus.resize(K);
int i;
for(i=0; i<N; i++){
windows w;
w.id = i;
win.push(w);
}
int h,m,pro;
for(i=0; i<K; i++){
scanf("%d",&pro);
cus[i].pro = pro;
windows w = win.top();
win.pop();
h=w.h; m=w.m;
if(w.c.size()==0){
//第一個客戶進入窗口
w.m = (m+pro)%60;
w.h = h + (m+pro)/60;
w.c.push(i);
}else if(w.c.size()>0 && w.c.size()<M){
w.c.push(i);
}else if(w.c.size()==M){
cus[w.c.front()].h = w.h;
cus[w.c.front()].m = w.m;
w.c.pop();
w.c.push(i);
int firstCus = w.c.front();
pro = cus[firstCus].pro;
w.m = (m+pro)%60;
w.h = h + (m+pro)/60;
}
win.push(w);
}
while(!win.empty()) {
windows w = win.top();
win.pop();
while(!w.c.empty()){
int index = w.c.front();
cus[index].h = w.h;
cus[index].m = w.m;
w.c.pop();
if(!w.c.empty()){
h = w.h;
m = w.m;
pro = cus[w.c.front()].pro;
w.m = (m + pro)%60;
w.h = h + (m + pro)/60;
}
}
}
int q;
for(i=0; i<Q; i++){
scanf("%d",&q);
h = cus[q-1].h;
m = cus[q-1].m;
if((h*60+m-cus[q-1].pro)>=(17*60)){
printf("Sorry\n");
}else{
printf("%02d:%02d\n",h,m);
}
}
return 0;
}