1095 Cars on Campus (30point(s)) - C语言 PAT 甲级

1095 Cars on Campus (30point(s))

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：

输入 N 个车牌号，时间，进出记录，查询 K 个时间点，输出校园内车辆的数目，最后并输出在校园内停留时间最长的车牌号和停留时间，若有多辆，车牌号按字典序从小到大输出

一辆车出入的配对要求：

• 多次连续进入，取最后一个值，
• 多次连续出来，取第一个值

设计思路：

按照车牌号字典序从小到大，时间从小到大对所有车进行排序

• 遍历且仅记录满足配对要求的每对记录
• 遍历过程中记录每辆车停留的时间
  • 同一辆车会有多次满足配对要求的记录，所以对于停留时间要判断累加
  • 并记录最大值和车牌号
• 利用满足配对要求的记录，按秒记录每个时间点在校园内的车辆数目，查询时，直接返回对应时间的数目即可

编译器：C (gcc)

#include 
#include 
#include 

struct node {
        char id[10];
        int time;
        int state;
};

int cmp1(const void *a, const void *b)
{
        struct node *x = (struct node *)a, *y = (struct node *)b;
        if (strcmp(x->id, y->id) != 0)
                return strcmp(x->id, y->id);
        return x->time - y->time;
}

int cmp2(const void *a, const void *b)
{
        return (*((struct node **)a))->time - (*((struct node **)b))->time;
}

int main(void)
{
        int n, k;
        int i, j, hh, mm, ss, tempint;
        struct node record[10000], *recordtime[10000];
        char tempstr[5], maxid[10000][10];
        int sum[24 * 3600], cnt, maxtime, maxcnt;

        scanf("%d%d\n", &n, &k);
        for (i = 0; i < n; i++) {
                scanf("%s %d:%d:%d %s\n", record[i].id, &hh, &mm, &ss, tempstr);
                tempint = hh * 3600 + mm * 60 + ss;
                record[i].time = tempint;
                record[i].state = tempstr[0] == 'i' ? 1 : -1;
        }
        qsort(record, n, sizeof(record[0]), cmp1);
        for (i = 1, j = 0; i < n; i++) {
                if (strcmp(record[i - 1].id, record[i].id) == 0 && record[i - 1].state == 1 && record[i].state == -1) {
                        record[j] = record[i - 1];
                        recordtime[j] = &record[j];
                        j++;
                        record[j] = record[i];
                        recordtime[j] = &record[j];
                        j++;

                        if (j == 2) {
                                tempint = record[j - 1].time - record[j - 2].time;
                                maxtime = tempint;
                                strcpy(maxid[0], record[j - 1].id);
                                maxcnt = 1;
                        } else if (j > 2) {
                                if (strcmp(record[j - 1].id, record[j - 3].id) == 0) {
                                        tempint += record[j - 1].time - record[j - 2].time;
                                } else {
                                        tempint = record[j - 1].time - record[j - 2].time;
                                }
                                if (maxtime < tempint) {
                                        maxtime = tempint;
                                        strcpy(maxid[0], record[j - 1].id);
                                        maxcnt = 1;
                                } else if (maxtime == tempint) {
                                        strcpy(maxid[maxcnt], record[j - 1].id);
                                        maxcnt++;
                                }
                        }
                }
        }
        n = j;
        qsort(recordtime, n, sizeof(recordtime[0]), cmp2);
        for (i = 0, j = 0, cnt = 0; i < 24 * 3600; i++) {
                while (j < n && recordtime[j]->time <= i) {
                        if (recordtime[j]->state == 1)
                                cnt++;
                        else
                                cnt--;
                        j++;
                }
                sum[i] = cnt;
        }

        for (i = 0; i < k; i++) {
                scanf("%d:%d:%d", &hh, &mm, &ss);
                tempint = hh * 3600 + mm * 60 + ss;
                printf("%d\n", sum[tempint]);
        }
        for (i = 0; i < maxcnt; i++)
                printf("%s ", maxid[i]);
        printf("%02d:%02d:%02d", maxtime / 3600, (maxtime % 3600) / 60, maxtime % 60);

        return 0;
}