1026 Table Tennis (30)（30 分）
A table tennis club has N tables available to the public. The tables are numbered from 1 to N. For any pair of players, if there are some tables open when they arrive, they will be assigned to the available table with the smallest number. If all the tables are occupied, they will have to wait in a queue. It is assumed that every pair of players can play for at most 2 hours.
Your job is to count for everyone in queue their waiting time, and for each table the number of players it has served for the day.
One thing that makes this procedure a bit complicated is that the club reserves some tables for their VIP members. When a VIP table is open, the first VIP pair in the queue will have the priviledge to take it. However, if there is no VIP in the queue, the next pair of players can take it. On the other hand, if when it is the turn of a VIP pair, yet no VIP table is available, they can be assigned as any ordinary players.
Input Specification:
Each input file contains one test case. For each case, the first line contains an integer N (<=10000) - the total number of pairs of players. Then N lines follow, each contains 2 times and a VIP tag: HH:MM:SS - the arriving time, P - the playing time in minutes of a pair of players, and tag - which is 1 if they hold a VIP card, or 0 if not. It is guaranteed that the arriving time is between 08:00:00 and 21:00:00 while the club is open. It is assumed that no two customers arrives at the same time. Following the players' info, there are 2 positive integers: K (<=100) - the number of tables, and M (< K) - the number of VIP tables. The last line contains M table numbers.
Output Specification:
For each test case, first print the arriving time, serving time and the waiting time for each pair of players in the format shown by the sample. Then print in a line the number of players served by each table. Notice that the output must be listed in chronological order of the serving time. The waiting time must be rounded up to an integer minute(s). If one cannot get a table before the closing time, their information must NOT be printed.
Sample Input:
9
20:52:00 10 0
08:00:00 20 0
08:02:00 30 0
20:51:00 10 0
08:10:00 5 0
08:12:00 10 1
20:50:00 10 0
08:01:30 15 1
20:53:00 10 1
3 1
2Sample Output:
08:00:00 08:00:00 0
08:01:30 08:01:30 0
08:02:00 08:02:00 0
08:12:00 08:16:30 5
08:10:00 08:20:00 10
20:50:00 20:50:00 0
20:51:00 20:51:00 0
20:52:00 20:52:00 0
3 3 2我的代码：
#include <stdio.h>
#include<iostream>
#include <algorithm>
using namespace std; struct Play{
int arrive,serve,ing,ed,vip;
int waiting;
Play(){
serve=-;//表示没被服务。
}
}play[];
struct Table{
int vip,people;
int ing;//是否正在使用
Table(){
vip=;ing=-;people=;
}
}table[];
bool cmp(Play& a,Play& b){
return a.arrive<b.arrive;//从小到大。
}
bool cmp2(Play& a,Play& b){
return a.serve<b.serve;
}
int main() {
int n,m,vm,vmno;
scanf("%d",&n);
int h,miu,s,bg=*;
for(int i=;i<n;i++){
scanf("%d:%d:%d",&h,&miu,&s);
play[i].arrive=h*+miu*+s-bg;
scanf("%d %d",&play[i].ing,&play[i].vip);
play[i].ing*=;
if(play[i].ing>)
play[i].ing=;
}
sort(play,play+n,cmp);
scanf("%d%d",&m,&vm);
for(int i=;i<=vm;i++){
scanf("%d",&vmno);
table[vmno].vip=;//是vip桌子。
}
// for(int i=0;i<n;i++){
// printf("%02d:%02d:%02d ",play[i].arrive/3600+8,play[i].arrive%3600/60,play[i].arrive%60);
// printf("%d %d\n",play[i].ing,play[i].vip);
// //printf("%02d:%02d:%02d ",play[i].serve/3600+8,play[i].serve%3600/60,play[i].serve%60);
// } //一定要注意要服务一个人的时候，需要保证那个人已经到了。。。
//注意，题目要求，如果当前队列中没有VIP选手，那么正常选手可以使用VIP桌子。
int ct=;
for(int tm=;tm<;tm++){
//判断当前是否有结束的
//不能用来计数，因为有可能serve到。
if(ct==n-)break;//人已经分配完了。
for(int i=;i<=m;i++){
if(table[i].ing!=-&&play[table[i].ing].ed==tm){//判断当前服务是否结束
table[i].ing=-;
}
}
int bg=,j,k;//记录开始循环的地方。
for(int i=;i<=m;i++){
j=bg;
if(table[i].ing==-&&table[i].vip==){//分配空闲vip桌子,分配vip用户。
for(j=bg;j<n;j++){
if(play[j].arrive<=tm&&play[j].vip==&&play[j].serve==-){//当前选手是vip,且没被服务
table[i].ing=j;
table[i].people++;
play[j].serve=tm;
play[j].ed=tm+play[j].ing;ct++;
// printf("%d %d %d %d vip\n",i,j,tm,tm-play[j].arrive);
break;//跳出内层循环
}
}
bg=j;
}
}
bg=;
for(int i=;i<=m;i++){
k=bg;
if(table[i].ing==-){//分配桌子与群众
for(k=bg;k<n;k++){
if(play[k].arrive<=tm&&play[k].serve==-){
table[i].ing=k;
table[i].people++;
play[k].serve=tm;
play[k].ed=tm+play[k].ing;ct++;
// printf("%d %d %d %d\n",i,j,tm,tm-play[j].arrive);
break;
}
}
bg=k;//加了这个之后就可以了，不用每次都从头开始遍历。有一个标记。
}
} }
ct=;
sort(play,play+n,cmp2);
for(int i=;i<n;i++){
if(play[i].serve!=-){
printf("%02d:%02d:%02d ",play[i].arrive/+,play[i].arrive%/,play[i].arrive%);
printf("%02d:%02d:%02d ",play[i].serve/+,play[i].serve%/,play[i].serve%);
if((play[i].serve-play[i].arrive)%>=)
ct=;
else
ct=;
printf("%d\n",(play[i].serve-play[i].arrive)%/+ct);
}
}
printf("%d",table[].people);
for(int i=;i<=m;i++){
printf(" %d",table[i].people);
}
return ;
}
//！！！在牛客网上AC 了，但是在pat上就只有16分，只通过了两个点，其他的都是答案错误，真是哭唧唧，完全不知道该怎么改了。！！先放这。明天复习一下。绝望了。
其中第一次提交通过80%，有如下没通过：
测试用例:
4
08:00:00 60 0
08:10:00 50 0
08:50:00 30 0
09:00:00 30 1
2 1
2
对应输出应该为:
08:00:00 08:00:00 0
08:10:00 08:10:00 0
09:00:00 09:00:00 0
08:50:00 09:00:00 10
2 2
你的输出为:
08:00:00 08:00:00 0
08:10:00 08:10:00 0
08:50:00 09:00:00 10
改了cmp2函数之后，提交之后报：您的程序未能在规定时间内运行结束，请检查是否循环有错或算法复杂度过大。
看来我这么循环着做，不太对啊。复杂度太高了，一直循环。
大佬的代码：https://www.liuchuo.net/archives/2955
#include <iostream>
#include <vector>
#include <algorithm>
#include <cmath>
using namespace std;
struct person {
int arrive, start, time;
bool vip;
}tempperson;
struct tablenode {
int end = * , num;
bool vip;
};
bool cmp1(person a, person b) {
return a.arrive < b.arrive;
}
bool cmp2(person a, person b) {
return a.start < b.start;
}
vector<person> player;
vector<tablenode> table;
void alloctable(int personid, int tableid) {
if(player[personid].arrive <= table[tableid].end)
player[personid].start = table[tableid].end;
//如果客户已经到了，当前vip桌子还没结束，那么开始服务时间就被设置为这个桌子的结束时间
//这个不按照时间线过来，还是有点颠覆我以前的做法了。
else
player[personid].start = player[personid].arrive;
table[tableid].end = player[personid].start + player[personid].time;
//当前桌子的结束时间被设置为客户开始服务时间+客户需服务时间
table[tableid].num++;
}
int findnextvip(int vipid) {
vipid++;
while(vipid < player.size() && player[vipid].vip == false) {
vipid++;
}
return vipid;
}
int main() {
int n, k, m, viptable;
scanf("%d", &n);
for(int i = ; i < n; i++) {
int h, m, s, temptime, flag;
scanf("%d:%d:%d %d %d", &h, &m, &s, &temptime, &flag);
tempperson.arrive = h * + m * + s;//转换成秒
tempperson.start = * ;//所有的开始服务时间都被设置为最终点。
if(tempperson.arrive >= * )
continue;
tempperson.time = temptime <= ? temptime * : ;//限制时间。
tempperson.vip = ((flag == ) ? true : false);
player.push_back(tempperson);
}
scanf("%d%d", &k, &m);
table.resize(k + );
for(int i = ; i < m; i++) {
scanf("%d", &viptable);
table[viptable].vip = true;
}
sort(player.begin(), player.end(), cmp1);//对向量可以这么排序。
int i = , vipid = -;
vipid = findnextvip(vipid);//找到队列中vip所在的下标
while(i < player.size()) {//这个是基于客户来循环。
int index = -, minendtime = ;
for(int j = ; j <= k; j++) {//k表示窗口的数量
if(table[j].end < minendtime) {
minendtime = table[j].end;//对每个桌子标记了结束时间。
index = j;//找到最近时间空闲的窗口
}
}
if(table[index].end >= * )
break;//如果所有桌子的最小服务结束时间>=21*3600，那么就结束
if(player[i].vip == true && i < vipid) {
i++;
continue;
}
if(table[index].vip == true) {
if(player[i].vip == true) {//如果当前客户i是vip客户。
alloctable(i, index);
if(vipid == i)
vipid = findnextvip(vipid);
i++;
} else {
if(vipid < player.size() && player[vipid].arrive <= table[index].end) {
//如果当前vipid不超过队列大小，并且在这张桌子使用结束前到来。
alloctable(vipid, index);
vipid = findnextvip(vipid);//寻找下一个vip
} else {
alloctable(i, index);//这个队列中没有vip了，那么就顺位。
i++;
}
}
} else {//不是vip桌子。
if(player[i].vip == false) {
alloctable(i, index);
i++;
} else {//如果当前是vip客户，但不是vip桌子
int vipindex = -, minvipendtime = ;
for(int j = ; j <= k; j++) {//遍历所有桌子，找出使用距离最近的桌子
if(table[j].vip == true && table[j].end < minvipendtime) {
minvipendtime = table[j].end;
vipindex = j;
}
}
if(vipindex != - && player[i].arrive >= table[vipindex].end) {
alloctable(i, vipindex);
if(vipid == i)
vipid = findnextvip(vipid);//找下一个。
i++;
} else {
alloctable(i, index);
if(vipid == i)
vipid = findnextvip(vipid);
i++;
}
}
}
}
sort(player.begin(), player.end(), cmp2);
for(i = ; i < player.size() && player[i].start < * ; i++) {
printf("%02d:%02d:%02d ", player[i].arrive / , player[i].arrive % / , player[i].arrive % );
printf("%02d:%02d:%02d ", player[i].start / , player[i].start % / , player[i].start % );
printf("%.0f\n", round((player[i].start - player[i].arrive) / 60.0));
}
for(int i = ; i <= k; i++) {
if(i != )
printf(" ");
printf("%d", table[i].num);
}
return ;
}
//这题真的太难了，看晕了，暂时放弃。
2018-11-22更————
今天又看了以下，还是没发现问题，不知道问题出在哪里，...太难了这道，暂时放弃。