Door opener and closing man

Door opener and closing man

Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)

Total submission (s): 3451 Accepted Submission (s): 1646

Problem description The first person to the computer room every day to open the door, the last one to leave the door closed. There are a lot of cluttered room to sign, check out records, please according to the records to find the day open and close the door.

The first line of input test inputs gives the total number of days of records N (> 0). The following is a list of N-Day records. Each day the record in the first row gives the number of entries in the record m (> 0), the following is the M row, each line is formatted as Passport number check-in time sign-off time The time is given by "hours: minutes: Seconds" (2 bits each), and the ID number is a string of no more than 15 length.

Output 1 lines for each day's record, that is, the opening and closing of the person's identification number, in the middle with 1 spaces separated. Note: In the referee's standard test input, all records are guaranteed to be complete, and each person's check-in time is before the time of checkout, and there are not many people who sign in or sign out at once.

Sample Input 31me3021112225321 00:00:00 23:59:592ee301218 08:05:35 20:56:35ma301134 12:35:45 21:40:423cs301111 15:30:28 17:00:10sc3021234 08:00:00 11:25:25cs301133 21:45:00 21:58:40

Sample Output ME3021112225321 me3021112225321ee301218 ma301134sc3021234 CS301133

Direct sort, the water

# include <iostream># include<cstdio># include<string># include<algorithm># include<vector>using namespacestd;structpeople{stringname; stringJin; stringChu;};BOOLCmp1 (People A, people b) {returnA.jin <B.jin;}BOOLCMP2 (People A, people b) {returnA.chu >B.chu;}intMain () {intT; CIN>>T;  while(t--) {vector<People>v;        People p; intN; CIN>>N;  for(inti =0; I < n; i++) {cin>> p.name >> P.jin >>P.chu;        V.push_back (P);            } sort (V.begin (), V.end (), CMP1); cout<< v[0].name <<" ";            Sort (V.begin (), V.end (), CMP2); cout<< v[0].name <<Endl; }    return 0;}

Door opener and closing man