题干：
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.
Input Specification:
Each input file contains one test case. Each case starts with two positive integers N (≤104), the number of records, and K (≤8×104) 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 ascending order of the times.
Output Specification:
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.
Sample Input:
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
Sample Output:
1
4
5
2
1
0
1
JH007BD ZD00001 07:20:09
题目大意：
给出n个车牌号、时间点、进出状态的记录，然后查询k个时间点这时校园内的车辆个数。最后还要输出在校园里面呆的时间最长的车的车牌号，以及呆了多久的时间。如果有多辆车就按照它的字母从小到大输出车牌。
配对要求是，如果一个车多次进入未出，取最后一个值；如果一个车多次out未进入，取第一个值。
注意：一个车可能出入校园好多次，停车的时间应该取之和。
解题报告：
注意凑样例的过程中，发现是要先对记录表单进行排错，然后再进行统计。
AC代码：
using namespace std;
typedef pair<int,int> PII;
const int MAX = 2e5 + 5;
char s[105],op[105];
int n,q;
struct Node {
string name;
int op;
int Time;
} R[MAX];
bool cmp(Node a,Node b) {
if(a.Time != b.Time) return a.Time < b.Time;
else return a.op < b.op;
}
map<string,int> state;//记录每辆车的状态
map<string,int> Time;//每辆车的停车时间
map<string,int> has;
vector<string> ans;
int anst;
bool ok[MAX];
int main()
{
memset(ok,1,sizeof ok);
cin>>n>>q;
for(int hh,mm,ss,i = 1; i<=n; i++) {
scanf("%s",s);
scanf("%d:%d:%d",&hh,&mm,&ss);
scanf("%s",op);
R[i].name = s;
R[i].op = (op[0] == 'i' ? 1 : 2);
R[i].Time = hh*3600+mm*60+ss;
}
for(int hh,mm,ss,i = 1; i<=q; i++) {
scanf("%d:%d:%d",&hh,&mm,&ss);
R[n+i].op = 3;
R[n+i].Time = hh*3600+mm*60+ss;
}
sort(R+1,R+n+q+1,cmp);
has.clear();
for(int i = 1; i<=n+q; i++) {
if(R[i].op == 1) {
has[R[i].name] = 1;
}
else if(R[i].op == 2){
if(has[R[i].name] == 0) ok[i] = 0;
has[R[i].name] = 0;
}
}
for(int i = 1; i<=n+q; i++) has[R[i].name] = -1;
for(int i = n+q; i>=1; i--) {
if(ok[i] == 0) continue;
if(R[i].op == 2) {
has[R[i].name] = 0;
}
else if(R[i].op == 1) {
if(has[R[i].name] != 0) ok[i] = 0;
has[R[i].name] = 1;
}
}
has.clear();
int cur = 0;
for(int i = 1; i<=n+q; i++) {
if(ok[i] == 0) continue;
if(R[i].op == 3) printf("%d\n",cur);
else if(R[i].op == 1) {
state[R[i].name] = R[i].Time;
has[R[i].name] = 1;
cur++;
}
else {
has[R[i].name] = 0;
cur--;
Time[R[i].name] += R[i].Time - state[R[i].name];
}
}
for(auto tar : Time) {
if(tar.second > anst) {
anst = tar.second; ans.clear(); ans.pb(tar.first);
}
else if(tar.second == anst) {
ans.pb(tar.first);
}
}
sort(ans.begin(),ans.end());
for(auto tar : ans) {
cout << tar << " ";
}
int hh,mm,ss;
hh = anst / 3600; anst -= hh * 3600;
mm = anst / 60; anst -= mm * 60;
ss = anst;
printf("%02d:%02d:%02d\n",hh,mm,ss);
return 0 ;
}
