inperson

sort排序问题

问题

1006SignInandSignOut(25分)

Atthebeginningofeveryday,thefirstpersonwhosignsinthecomputerroomwillunlockthedoor,andthelastonewho

herecordsofsigningin’sandout’s,youaresuppodtofindtheoneswhohave

unlockedandlockedthedooronthatday.

InputSpecification:

estartswithapositiveintegerM,

whichisthetotalnumberofrecords,followedbyMlines,eachintheformat:

ID_numberSign_in_timeSign_out_time

wheretimesaregivenintheformatHH:MM:SS,andID_numberisastringwithnomorethan15characters.

OutputSpecification:

Foreachtestca,outputinonelinetheIDnumbersofthepersonswhohaveunlockedandlockedthedooronthatday.

ThetwoIDnumbersmustbeparatedbyonespace.

Note:,thesignintimemustbeearlierthanthesignouttimefor

eachperson,andtherearenotwopersonssigninoroutatthesamemoment.

SampleInput:

3

CS30111115:30:2817:00:10

SC302123408:00:0011:25:25

CS30113321:45:0021:58:40

结尾⽆空⾏

SampleOutput:

SC3021234CS301133

结尾⽆空⾏

思路

利⽤sort函数对两个结构体数组按照不同的规则排序。对于in数组，按照⼩时，分钟，秒的权重从⼩到⼤排序，最⼩的元素即为最早签到的

⼈。对于out数组，按照⼩时，分钟，秒的权重从⼤到⼩排序，最⼩的元素即为最晚签离的⼈。

代码

#include

usingnamespacestd;

constintmaxn=1010;

structnode{

stringid;

intHH,MM,SS;

node(){}

node(string_id,int_HH,int_MM,int_SS):id(_id),HH(_HH),MM(_MM),SS(_SS){}

};

structnodein[maxn],out[maxn];

intm;

boolcmpIn(nodea,nodeb){

if(!=)

<;

el{

if(!=)

<;

el

<;

}

}

boolcmpOut(nodea,nodeb){

if(!=)

>;

el{

if(!=)

>;

el

>;

}

}

intmain(){

intH,M,S,H1,M1,S1;

cin>>m;

stringid;

for(inti=0;i

cin>>id;

getchar();

scanf("%d:%d:%d",&H,&M,&S);

getchar();

scanf("%d:%d:%d",&H1,&M1,&S1);

nodeinPerson=node(id,H,M,S);

in[i]=inPerson;

nodeoutPerson=node(id,H1,M1,S1);

out[i]=outPerson;

}

sort(in,in+m,cmpIn);

sort(out,out+m,cmpOut);

cout<

return0;

}

总结

⽤上结构体的构造函数来简化代码。