PAT 1095 Cars on Campus-白红宇

PAT 1095 Cars on Campus

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发布时间：2019-06-12

本文共 5042 字，大约阅读时间需要 16 分钟。

1095 Cars on Campus （30 分）

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

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

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 7JH007BD 18:00:01 inZD00001 11:30:08 outDB8888A 13:00:00 outZA3Q625 23:59:50 outZA133CH 10:23:00 inZD00001 04:09:59 inJH007BD 05:09:59 inZA3Q625 11:42:01 outJH007BD 05:10:33 inZA3Q625 06:30:50 inJH007BD 12:23:42 outZA3Q625 23:55:00 inJH007BD 12:24:23 outZA133CH 17:11:22 outJH007BD 18:07:01 outDB8888A 06:30:50 in05:10:0006:30:5011:00:0012:23:4214:00:0018:00:0023:59:00

Sample Output:

1452101JH007BD ZD00001 07:20:09

题目大意：需要统计在特定的时间在校园内停车的总数，并且找出本天停车时间最长的那辆车。输入数据有n个车辆进出信息，并有m个时间点查询。

每个车辆有车牌号时间状态，并且每个in都和下一个最近的out匹配，没有匹配的in或者out忽略掉。给定的查询时间是递增的。

//将时间都转换为int存储。每来一个就查询一次时间复杂度好高啊，8次方了，千万级别。

//但是如果有两个in，那么谁和第一个out对呢？那肯定是最后一个。

//思路错了，如果有一辆车in out in out你这种方法就不行了。

#include 
      
       #include 
       
        #include 
        #include 
         
          #include
          
           using namespace std;struct Car{    char name[10];    int in,out;    Car(char *n,int i,int o){        name=n;in=i;out=o;//这样易于后边的排出。    }};map
           
             id2name;map
            
              name2id;vector
             
               cars;int main() { int n,m; scanf("%d%d",&m,&n); char pl[10],state[5]; int h,m,s; string nm; for(int i=0;i

View Code

//写到这里写不下去了，不知道该如何处理，如何判断。

代码转自：https://www.liuchuo.net/archives/2951

#include 
     
      #include 
      
       #include 
       
        #include 
        
         #include 
         
          #include 
          using namespace std;struct node { char id[10]; int time, flag = 0;};bool cmp1(node a, node b) { if(strcmp(a.id, b.id) != 0) return strcmp(a.id, b.id) < 0;//对所有的输入按照id排序。 else return a.time < b.time;//同一辆车时间小的在前。}bool cmp2(node a, node b) { return a.time < b.time;}int main() { int n, k, maxtime = -1, tempindex = 0; scanf("%d%d\n", &n, &k); vector
           
             record(n), car;//将其确定大小，后序可以使用下标访问。 for(int i = 0; i < n; i++) { char temp[5]; int h, m, s; //输入这个字符串的时候不用再&符号了！ scanf("%s %d:%d:%d %s\n", record[i].id, &h, &m, &s, temp); int temptime = h * 3600 + m * 60 + s; record[i].time = temptime; record[i].flag = strcmp(temp, "in") == 0 ? 1 : -1; } sort(record.begin(), record.end(), cmp1); map
            
              mapp;//id映射到时间。 for(int i = 0; i < n - 1; i++) { if(strcmp(record[i].id, record[i+1].id) == 0 && record[i].flag == 1 && record[i+1].flag == -1) { car.push_back(record[i]); car.push_back(record[i+1]); mapp[record[i].id] += (record[i+1].time - record[i].time); if(maxtime < mapp[record[i].id]) { maxtime = mapp[record[i].id]; } } } sort(car.begin(), car.end(), cmp2);//再对car进行排序。 vector
             
               cnt(n); for(int i = 0; i < car.size(); i++) { if(i == 0) cnt[i] += car[i].flag; else cnt[i] = cnt[i - 1] + car[i].flag; } for(int i = 0; i < k; i++) { int h, m, s; scanf("%d:%d:%d", &h, &m, &s); int temptime = h * 3600 + m * 60 + s; int j; for(j = tempindex; j < car.size(); j++) { if(car[j].time > temptime) { //如果时间已经大于了当前，那么就不算。 printf("%d\n", cnt[j - 1]); break; } else if(j == car.size() - 1) { printf("%d\n", cnt[j]); } } tempindex = j; } for(map
              
               ::iterator it = mapp.begin(); it != mapp.end(); it++) { if(it->second == maxtime) printf("%s ", it->first.c_str()); } printf("%02d:%02d:%02d", maxtime / 3600, (maxtime % 3600) / 60, maxtime % 60); return 0;}

//这个cnt数组思路很好，学习了，需要复习啊，待会回想一下。

转载于:https://www.cnblogs.com/BlueBlueSea/p/9801488.html

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