地址:
http://ac.jobdu.com/problem.php?pid=1013
題目描述:
每天第一個到機房的人要把門打開,最後一個離開的人要把門關好。現有一堆雜亂的機房簽到、籤離記錄,請根據記錄找出當天開門和關門的人。
輸入:
測試輸入的第一行給出記錄的總天數N ( N> 0 ),下面列出了N天的記錄。
每天的記錄在第一行給出記錄的條目數M (M > 0 ),下面是M行,每行的格式爲
證件號碼 簽到時間 籤離時間
其中時間按“小時:分鐘:秒鐘”(各佔2位)給出,證件號碼是長度不超過15的字符串。
輸出:
對每一天的記錄輸出1行,即當天開門和關門人的證件號碼,中間用1空格分隔。
注意:在裁判的標準測試輸入中,所有記錄保證完整,每個人的簽到時間在籤離時間之前,且沒有多人同時簽到或者籤離的情況。
樣例輸入:
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
樣例輸出:
ME3021112225321 ME3021112225321
EE301218 MA301134
SC3021234 CS301133
來源:
2005年浙江大學計算機及軟件工程研究生機試真題
源碼:
#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
int n, m;
char record[ 40 ]; //記錄
struct WORKER{
char name[ 20 ];
int startH; //到達的小時
int startM; //到達的分鐘
int startS; //到達的秒鐘
int endH; //離開的小時
int endM; //離開的分鐘
int endS; //離開的秒鐘
}worker[100];
int workerNum = 0;
//將記錄信息轉化到結構體存儲
void recordToWorker(){
int id = 0;
while( record[ id ] != ' '){
worker[ workerNum ].name[ id ] = record[ id ];
id++;
}
worker[ workerNum ].name[ id ] = '\0';
id ++; //跳過空格
int a, b;
a = ( int )( record[ id++ ] - '0' );
b = ( int )( record[ id++ ] - '0' );
worker[ workerNum ].startH = a * 10 + b;
id ++; //跳過冒號
a = ( int )( record[ id++ ] - '0' );
b = ( int )( record[ id++ ] - '0' );
worker[ workerNum ].startM = a * 10 + b;
id ++; //跳過冒號
a = ( int )( record[ id++ ] - '0' );
b = ( int )( record[ id++ ] - '0' );
worker[ workerNum ].startS = a * 10 + b;
id ++; //跳過空格
a = ( int )( record[ id++ ] - '0' );
b = ( int )( record[ id++ ] - '0' );
worker[ workerNum ].endH = a * 10 + b;
id ++; //跳過冒號
a = ( int )( record[ id++ ] - '0' );
b = ( int )( record[ id++ ] - '0' );
worker[ workerNum ].endM = a * 10 + b;
id ++; //跳過冒號
a = ( int )( record[ id++ ] - '0' );
b = ( int )( record[ id++ ] - '0' );
worker[ workerNum ].endS = a * 10 + b;
workerNum++;
}
//按照到達時間排序
bool cmpStart( WORKER wor1, WORKER wor2 ){
if( wor1.startH != wor2.startH ){
return wor1.startH < wor2.startH;
}
else if( wor1.startM != wor2.startM ){
return wor1.startM < wor2.startM;
}
else{
return wor1.startS < wor2.startS;
}
}
//按照離開時間排序
bool cmpEnd( WORKER wor1, WORKER wor2 ){
if( wor1.endH != wor2.endH ){
return wor1.endH > wor2.endH;
}
else if( wor1.endM != wor2.endM ){
return wor1.endM > wor2.endM;
}
else{
return wor1.endS > wor2.endS;
}
}
int main(){
while( scanf("%d", &n) != EOF ){
while( n-- ){
scanf("%d", &m);
int idM = m;
workerNum = 0;
getchar();
//輸入數據
while( idM-- ){
gets( record );
recordToWorker();
}
sort( worker, worker+m, cmpStart);
for( int i = 0; worker[0].name[ i ] != '\0'; i ++ ){
printf( "%c", worker[0].name[i ] );
}
printf( " " );
sort( worker, worker+m, cmpEnd);
for( int i = 0; worker[0].name[ i ] != '\0'; i ++ ){
printf( "%c", worker[0].name[ i ] );
}
printf( "\n" );
}
}
}
/**************************************************************
Problem: 1013
User: 螺小旋
Language: C++
Result: Accepted
Time:0 ms
Memory:1032 kb
****************************************************************/