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Zhejiang University has 6 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 (<= 10000), the number of records, and K (<= 80000) 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 23:59:00Sample Output:
1 4 5 2 1 0 1 JH007BD ZD00001 07:20:09
将record按照时间顺序进行排序，先确定每个record是否paired（用一个map来记录某一个车牌进入的record，遇到相应的out的记录，将两个都设为paired）
然后计算每个时间点的车辆数，以及最后最大时长的车辆。
注意此题的query较多，不能用cin输入否则会超时，需要用scanf输入
1 #include <iostream> 2 #include <map> 3 #include <string> 4 #include <algorithm> 5 #include <vector> 6 #include <bitset> 7 #include <sstream> 8 9 using namespace std; 10 11 struct Record 12 { 13 string plateNum; 14 string time; 15 string status; 16 }; 17 18 Record records[10000]; 19 bitset<10000> paired; 20 map<string, int> in; 21 string query[80000]; 22 23 bool cmp(Record r1, Record r2) 24 { 25 return r1.time < r2.time; 26 } 27 28 int Duration(string start, string end) 29 { 30 for (int i = 0; i < start.size(); i++) 31 if (start[i] == ‘:‘) 32 start[i] = end[i] = ‘ ‘; 33 34 int h1, h2, m1, m2, s1, s2; 35 stringstream ss; 36 ss << start << " " << end; 37 ss >> h1 >> m1 >> s1 >> h2 >> m2 >> s2; 38 39 return (h2 * 60 * 60 + m2 * 60 + s2) - (h1 * 60 * 60 + m1 * 60 + s1); 40 } 41 42 string ToStandardTime(int duration) 43 { 44 int h, m, s; 45 h = duration / 3600; 46 m = (duration - 3600 * h) / 60; 47 s = duration - 3600 * h - m * 60; 48 49 return string(1, h / 10 + ‘0‘) + string(1, h % 10 + ‘0‘) + ":" + 50 string(1, m / 10 + ‘0‘) + string(1, m % 10 + ‘0‘) + ":" + 51 string(1, s / 10 + ‘0‘) + string(1, s % 10 + ‘0‘); 52 } 53 54 int main() 55 { 56 int recordNum, queryNum; 57 cin >> recordNum >> queryNum; 58 for (int i = 0; i < recordNum; i++) 59 cin >> records[i].plateNum >> records[i].time >> records[i].status; 60 for (int i = 0; i < queryNum; i++) 61 { 62 char s[10]; 63 scanf("%s", s); 64 query[i] = string(s); 65 } 66 67 sort(records, records + recordNum, cmp); 68 //sure whether a record is paired 69 for (int i = 0; i < recordNum; i++) 70 { 71 if (records[i].status == "in") 72 in[records[i].plateNum] = i; 73 else 74 { 75 map<string, int>::iterator it = in.find(records[i].plateNum); 76 if ( it != in.end()) 77 { 78 int indexOfin = in[records[i].plateNum]; 79 paired.set(indexOfin); 80 paired.set(i); 81 in.erase(it); 82 } 83 } 84 } 85 in.clear(); 86 87 int queryIndex = 0; 88 string querytime = query[queryIndex++]; 89 map<string, int> duration; 90 for (int i = 0; i < recordNum; i++) 91 { 92 while(querytime < records[i].time) 93 { 94 cout << in.size() << endl; 95 if (queryIndex < queryNum) 96 querytime = query[queryIndex++]; 97 else 98 querytime = "24:00:00"; 99 } 100 if (paired[i]) 101 { 102 if (records[i].status == "in") 103 in[records[i].plateNum] = i; 104 else 105 { 106 int indexOfIn = in[records[i].plateNum]; 107 if (duration.find(records[i].plateNum) == duration.end()) 108 duration[records[i].plateNum] = Duration(records[indexOfIn].time, records[i].time); 109 else 110 duration[records[i].plateNum] += Duration(records[indexOfIn].time, records[i].time); 111 in.erase(in.find(records[i].plateNum)); 112 } 113 } 114 } 115 if (querytime != "24:00:00") 116 { 117 for (int i = 0; i <= queryNum - queryIndex; i++) 118 cout << 0 << endl; 119 } 120 121 //sure the maximum period and corresponding plate numbers 122 int maxDuration = -1; 123 vector<string> plate; 124 for (map<string, int>::iterator it = duration.begin(); it != duration.end(); it++) 125 { 126 if (it->second > maxDuration) 127 { 128 maxDuration = it->second; 129 plate.clear(); 130 plate.push_back(it->first); 131 } 132 else if (it->second == maxDuration) 133 plate.push_back(it->first); 134 } 135 for (vector<string>::iterator it = plate.begin(); it != plate.end(); it++) 136 cout << *it << " "; 137 cout << ToStandardTime(maxDuration); 138 }
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原文地址：http://www.cnblogs.com/jackwang822/p/4756930.html
