当前时刻停车厂有多少车，然后一天下来停的最久的车是哪几辆，时间是多久。
注意：
in、out是有不匹配的情况的。
还有题目里面说一辆车不会同时进出。我理解成几辆车不会同时进出了。结果发现样例都有同时进的。
匹配完，其实停车时间就出来了。如果车几次进出，最后求的是总的时间。
我直接开了个3601* 24的数组，每一秒来统计车辆的进出情况，然后从前往后累加，就是每个时刻的车辆数量。
#include <iostream> #include <vector> #include <string> #include <cstring> #include <algorithm> #include <set> #include <map> #define MAX 3601*24 using namespace std; int n, k; struct NodeCar { string id; int tag; int time; }; vector <NodeCar> list; struct NodeSeg { int time; string id; NodeSeg() { time = 0; } }; NodeSeg ParkTime[MAX]; int Count[MAX]; set <string> CarId; map <string, int> id2int; map <int, string> int2id; bool cmp(NodeCar n1, NodeCar n2) { if (n1.id != n2.id) return n1.id < n2.id; else return n1.time < n2.time; } bool cmp2(NodeSeg s1, NodeSeg s2) { if (s1.time != s2.time) return s1.time > s2.time; else return s1.id < s2.id; } void TimePrint(int time) { int hh = time / 3600; int mm = time % 3600 / 60; int ss = time % 3600 % 60; printf("d:d:d\n", hh, mm, ss); } int main() { cin >> n >> k; string s, tag; int hh, mm, ss, t; memset(Count, 0, sizeof(Count)); NodeSeg tempSeg; NodeCar tempCar; for (int i = 0; i < n; i++) { cin >> tempCar.id; CarId.insert(tempCar.id); cin >> hh; cin.get(); cin >> mm; cin.get(); cin >> ss >> tag; t = 3600 * hh + 60 * mm + ss; tempCar.time = t; if (tag == "in") tempCar.tag = 1; else if (tag == "out") tempCar.tag = 2; list.push_back(tempCar); } sort(list.begin(), list.end(), cmp); set <string>::iterator it; int num = 0; for (it = CarId.begin(); it != CarId.end(); it++) { id2int[*it] = num; int2id[num++] = *it; } //In、out匹配 bool isIn = false; int pre, p; for (int i = 0; i < list.size(); i++) { if (!isIn && list[i].tag == 1) { isIn = true; pre = i; } else if (isIn && list[i].tag == 1) { pre = i; } else if (isIn && list[i].tag == 2 && list[i].id == list[pre].id) { Count[list[pre].time]++; Count[list[i].time]--; ParkTime[id2int[list[i].id]].time += (list[i].time - list[pre].time); ParkTime[id2int[list[i].id]].id = list[pre].id; isIn = false; } } int sum = 0; for (int i = 0; i < MAX; i++) { sum += Count[i]; Count[i] = sum; } sort(ParkTime, ParkTime + CarId.size(), cmp2); for (int i = 0; i < k; i++) { cin >> hh; cin.get(); cin >> mm; cin.get(); cin >> ss; t = 3600 * hh + 60 * mm + ss; cout << Count[t] << endl; } int max = ParkTime[0].time; for (int i = 0; i < CarId.size(); i++) { if (ParkTime[i].time == max) { cout << ParkTime[i].id << " "; } } TimePrint(max); return 0; }