f439: 10191 - Longest Nap

解題心得

這題不難，只是很多瑣碎的地方要弄。

基本上就是把開始跟結束時間轉換成分鐘，然後用陣列來記錄哪些時候有空或沒空，最後再從頭掃到尾找最長的空檔。因為是10:00開始，把所有轉換後的時間減10*60分鐘。

寫完後覺得寫很醜，如果用開始跟結束時間作為標記，不知道會不會好看一點。

程式碼

#include <iostream>
#include <string>
using namespace std;

int t[480] = { 0 };

int to_int(string s)
{
	int n = 0;
	n = stoi(s.substr(0, 2)) * 60 + stoi(s.substr(3, 2)) - 600;
	return n;
}

int main()
{
	int testcase, counter = 1;
	while (cin >> testcase)
	{
		cin.ignore();
		for (int i = 0; i < 480; i++) t[i] = 0;
		for (int days = 1; days <= testcase; days++)
		{
			string s, start, end;
			getline(cin, s);
			start = s.substr(0, 5);
			end = s.substr(6, 5);
			for (int i = to_int(start); i <= to_int(end); i++)
				t[i] = 1;

			//cout << to_int(start) << " " << to_int(end) << endl;
		}
		int prev = -1, count = 0, maxCount = 0, curTime = 0, time = 0;
		for (int i = 0; i < 480; i++)
		{
			if (t[i] == 0)
				count++;
			if ((t[i] == 1 && prev == 0) || i == 479)
			{
				if (curTime == 0) count--;
				if (count + 1 > maxCount)
				{
					maxCount = count + 1;
					time = curTime;
				}
				count = 0;
				//cout << "update: " << maxCount << endl;
			}
			if (t[i] == 0 && prev == 1)
				curTime = i - 1;
			prev = t[i];
		}
		cout << "Day #" << counter++ << ": the longest nap starts at " << (time + 600) / 60 << ":";
		if ((time + 600) % 60 < 10) cout << "0" << (time + 600) % 60 ;
		else cout << (time + 600) % 60;
		cout << " and will last for ";
		if (maxCount < 60) cout << maxCount << " minutes.\n";
		else cout << maxCount / 60 << " hours and " << maxCount % 60 << " minutes.\n";

	}
	return 0;
}
// 18*60 - 10*60 = 8*60 = 480