开门人和关门人
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 21918 Accepted Submission(s): 10988
Problem Description
每天第一个到机房的人要把门打开，最后一个离开的人要把门关好。现有一堆杂乱的机房签
到、签离记录，请根据记录找出当天开门和关门的人。
Input
测试输入的第一行给出记录的总天数N ( > 0 )。下面列出了N天的记录。
每天的记录在第一行给出记录的条目数M ( > 0 )，下面是M行，每行的格式为
证件号码 签到时间 签离时间
其中时间按“小时:分钟:秒钟”（各占2位）给出，证件号码是长度不超过15的字符串。
Output
对每一天的记录输出1行，即当天开门和关门人的证件号码，中间用1空格分隔。
注意：在裁判的标准测试输入中，所有记录保证完整，每个人的签到时间在签离时间之前，
且没有多人同时签到或者签离的情况。
Sample Input
3 1 ME3021112225321 00:00:00 23:59:59 2 EE301218 08:05:35 20:56:35 MA301134 12:35:45 21:40:42 3 CS301111 15:30:28 17:00:10 SC3021234 08:00:00 11:25:25 CS301133 21:45:00 21:58:40
Sample Output
ME3021112225321 ME3021112225321 EE301218 MA301134 SC3021234 CS301133
Source
边输入边比较时间，更新开门人和关门人。
代码如下：
#include<stdio.h> #include<string.h> struct stu { int h; int m; int s; char name[20]; }open,close; int main() { int T; scanf("%d",&T); while(T--) { int n,i,oh,om,os,ch,cm,cs; char name[20]; scanf("%d",&n); for(i=0;i<n;i++) { scanf("%s %d:%d:%d %d:%d:%d",name,&oh,&om,&os,&ch,&cm,&cs); if(i==0) { open.h=oh; open.m=om; open.s=os; strcpy(open.name,name); close.h=ch; close.m=cm; close.s=cs; strcpy(close.name,name); } else { if(open.h>oh) { open.h=oh; open.m=om; open.s=os; strcpy(open.name,name); } else if(open.h==oh) { if(open.m==om&&open.s>os) { open.h=oh; open.m=om; open.s=os; strcpy(open.name,name); } else if(open.m>om) { open.h=oh; open.m=om; open.s=os; strcpy(open.name,name); } } if(close.h<ch) { close.h=ch; close.m=cm; close.s=cs; strcpy(close.name,name); } else if(close.h==ch) { if(close.m==cm&&close.s<cs) { close.h=ch; close.m=cm; close.s=cs; strcpy(close.name,name); } else if(close.m<cm) { close.h=ch; close.m=cm; close.s=cs; strcpy(close.name,name); } } } } printf("%s %s\n",open.name,close.name); } return 0; }
