Week14 限时大模拟 A - 猫睡觉问题

//如果时间段是跨夜的,那么结束时间的数值一定比开始时间小
//Exist==是否存在跨夜时间段,tot==活动的阶段数,totRest==时间大于等于A的空闲段的个数
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring> 
using namespace std;
int A, B, N;
int Exist, tot, totRest;
struct Active
{
	int beginH, beginM;
	int endH, endM;
	bool operator<(const Active& x) const
	{
		if (beginH != x.beginH) return beginH < x.beginH;
		else return beginM < x.beginM;
	}
}act[25], rest[50];

//返回两个时间段的差,以分钟计
int duration(int beginH, int beginM, int endH, int endM)
{
	if ( endH < beginH || ((endH == beginH) && (endM < beginM)) )
		return duration(beginH, beginM, 23, 59) + duration(0, 0, endH, endM);
	return  (endH - beginH) * 60 + (endM - beginM) + 1;		//+1是因为时间是闭区间 	
}

//统计空闲时间段,并找到大于等于A的段
void calc()
{
	for (int i = 1; i < tot; i++)
	{
		int len = duration(act[i].endH, act[i].endM, act[i + 1].beginH, act[i + 1].beginM);
		len = len - 2;
		if (len >= A)
			rest[++totRest] = { act[i].endH, act[i].endM, act[i + 1].beginH, act[i + 1].beginM };
	}
	if (Exist == 0)
	{
		//没有跨夜的时间段,就要把晚上最后的空闲和白天的第一个空闲连起来
		int len = duration(act[tot].endH, act[tot].endM, act[1].beginH, act[1].beginM);
		len = len - 2;
		if (len >= A)
			rest[++totRest] = { act[tot].endH, act[tot].endM, act[1].beginH, act[1].beginM };
	}
}

//判断是否满足条件:如果两个A段之间的时间大于B,则不可以 
bool Judge()
{
	if (totRest == 0) return false;//没有一个空闲时间段能休息 
	sort(rest + 1, rest + totRest + 1);
	//把第一个空闲时间段放在最后,为了组成一个环
	rest[++totRest] = rest[1];
	for (int i = 1; i < totRest; i++)
	{
		int len = duration(rest[i].endH, rest[i].endM, rest[i + 1].beginH, rest[i + 1].beginM);
		if (len > B) return false;
	}
	return true;
}
//输出方案 
void prit()
{
	cout << "Yes" << endl;
	cout << totRest - 1 << endl;
	for (int i = 1; i < totRest; i++)
	{
		//时间不能交叉 
		int b1 = rest[i].beginH, b2 = rest[i].beginM, e1 = rest[i].endH, e2 = rest[i].endM;
		//值得注意的是,分钟进位/借位的时候,要判断小时是不是23/0
		if (b2 == 59)
		{
			if (b1 == 23) b1 = 0;
			else b1 = b1 + 1;
			b2 = 0;
		}
		else b2 = b2 + 1;
		if (e2 == 0)
		{
			if (e1 == 0) e1 = 23;
			else e1 = e1 - 1;
			e2 = 59;
		}
		else e2 = e2 - 1;
		printf("%02d:%02d-%02d:%02d\n", b1, b2, e1, e2);
	}
}
int main()
{
	while (cin >> A >> B >> N)
	{
		Exist = 0, tot = 0, totRest = 0;
		memset(act, 0, sizeof(act));
		memset(rest, 0, sizeof(rest));
		A = A * 60; B = B * 60;
		for (int i = 1; i <= N; i++)
		{
			int b1, b2, e1, e2;
			scanf("%d:%d-%d:%d", &b1, &b2, &e1, &e2);
			if (b1 > e1)
			{
				//跨夜时间段,先分成两段处理
				Exist = 1;
				act[++tot] = { b1,b2,23,59 };
				act[++tot] = { 0,0,e1,e2 };
				continue;
			}
			act[++tot] = { b1,b2,e1,e2 };
		}
		sort(act + 1, act + tot + 1);
		calc();
		if (Judge()) prit();
		else cout << "No" << endl;
	}
	return 0;
}