Zhejiang University has 8 campuses and a lot of gates. From each gate we can collect the in/out times and the plate numbers of the cars crossing the gate. Now with all the information available, you are supposed to tell, at any specific time point, the number of cars parking on campus, and at the end of the day find the cars that have parked for the longest time period.
Each input file contains one test case. Each case starts with two positive integers N (≤), the number of records, and K (≤) the number of queries. Then N lines follow, each gives a record in the format:
plate_number hh:mm:ss status
where plate_number is a string of 7 English capital letters or 1-digit numbers; hh:mm:ss represents the time point in a day by hour:minute:second, with the earliest time being 00:00:00 and the latest 23:59:59; and status is either in or out.
Note that all times will be within a single day. Each in record is paired with the chronologically next record for the same car provided it is an out record. Any in records that are not paired with an out record are ignored, as are out records not paired with an in record. It is guaranteed that at least one car is well paired in the input, and no car is both in and out at the same moment. Times are recorded using a 24-hour clock.
Then K lines of queries follow, each gives a time point in the format hh:mm:ss. Note: the queries are given in accendingorder of the times.
For each query, output in a line the total number of cars parking on campus. The last line of output is supposed to give the plate number of the car that has parked for the longest time period, and the corresponding time length. If such a car is not unique, then output all of their plate numbers in a line in alphabetical order, separated by a space.
16 7
JH007BD 18:00:01 in
ZD00001 11:30:08 out
DB8888A 13:00:00 out
ZA3Q625 23:59:50 out
ZA133CH 10:23:00 in
ZD00001 04:09:59 in
JH007BD 05:09:59 in
ZA3Q625 11:42:01 out
JH007BD 05:10:33 in
ZA3Q625 06:30:50 in
JH007BD 12:23:42 out
ZA3Q625 23:55:00 in
JH007BD 12:24:23 out
ZA133CH 17:11:22 out
JH007BD 18:07:01 out
DB8888A 06:30:50 in
05:10:00
06:30:50
11:00:00
12:23:42
14:00:00
18:00:00
23:59:00
1 4 5 2 1 0 1 JH007BD ZD00001 07:20:09
注意点：1.查询时 now 必须放在外面不然会重复循环导致超时！！！
2.清空cout缓存用cout<<flush;否则使用ios::sync_with_stdio(false);
cin.tie(0)时不与printf兼容，可能输出顺序错误。
1 #include<bits/stdc++.h> 2 using namespace std; 3 4 const int maxn=10010; 5 6 struct Record{ 7 string carId; 8 int hh; 9 int mm; 10 int ss; 11 int time; 12 string status; 13 int flag; 14 }rec[maxn],valid[maxn]; 15 16 int n,k; 17 18 map<string,int> parkTime; 19 20 int num=0; 21 int maxTime=-1; 22 23 24 bool cmp1(Record a,Record b){ 25 if(a.carId!=b.carId) 26 return a.carId<b.carId; 27 else 28 return a.time<b.time; 29 } 30 31 bool cmp2(Record a,Record b){ 32 33 return a.time<b.time; 34 35 } 36 37 38 39 int main(){ 40 41 ios::sync_with_stdio(false); 42 cin.tie(0); 43 44 cin>>n>>k; 45 46 // scanf("%d%d",&n,&k); 47 48 49 50 char c; 51 Record temp; 52 53 for(int i=0;i<n;i++){ 54 cin>>temp.carId>>temp.hh>>c>>temp.mm>>c>>temp.ss>>temp.status; 55 56 //char str1[100]; 57 //char str2[100]; 58 //scanf("%s %d:%d:%d %s",&str1[0],&temp.hh,&temp.mm,&temp.ss,&str2[0]); 59 //temp.carId=str1; 60 //temp.status=str2; 61 62 63 temp.time=temp.hh*3600+temp.mm*60+temp.ss; 64 65 if(temp.status=="in") 66 temp.flag=1; 67 else 68 temp.flag=2; 69 70 rec[i]=temp; 71 } 72 73 74 sort(rec,rec+n,cmp1); 75 76 77 for(int i=0;i<n-1;i++){ 78 if(rec[i].carId==rec[i+1].carId&& 79 rec[i].flag==1&&rec[i+1].flag==2){ 80 valid[num++]=rec[i]; 81 valid[num++]=rec[i+1]; 82 83 string carId=rec[i].carId; 84 85 if(parkTime.count(carId)==0){ 86 parkTime[carId]=0; 87 } 88 89 parkTime[carId]+=(rec[i+1].time-rec[i].time); 90 91 maxTime=max(maxTime,parkTime[carId]); 92 93 } 94 } 95 96 97 98 sort(valid,valid+num,cmp2); 99 100 int now=0,nowCar=0; //now必须放外面 101 102 for(int i=0;i<k;i++){ 103 int hh,mm,ss; 104 char c; 105 106 cin>>hh>>c>>mm>>c>>ss; 107 108 //scanf("%d:%d:%d",&hh,&mm,&ss); 109 110 int queryTime=hh*3600+mm*60+ss; 111 112 113 114 115 while(now<num&&valid[now].time<=queryTime){ 116 if(valid[now].flag==1) nowCar++; 117 else nowCar--; 118 119 now++; 120 121 } 122 123 cout<<nowCar<<endl; 124 125 //printf("%d\n",nowCar); 126 127 } 128 129 130 131 132 for(map<string,int>::iterator it = parkTime.begin(); it != parkTime.end();it++){ 133 if(it->second==maxTime) 134 { 135 cout<<it->first<<" "; 136 137 // printf("%s ",it->first.c_str()); 138 } 139 } 140 141 cout<<flush; //清空cout缓存 142 143 //cout<<maxTime<<" "<<parkTime["JH007BD"]<<" "<<parkTime["ZD00001"]<<endl; 144 145 printf("%02d:%02d:%02d\n",maxTime/3600,(maxTime/60)%60,maxTime%60); 146 147 148 149 return 0; 150 }
原文：https://www.cnblogs.com/moranzju/p/11100082.html