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HDU 3700 Cat

編輯：C++入門知識

HDU 3700 Cat

Cat

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 451 Accepted Submission(s): 100
Special Judge

Problem Description There is a cat, cat likes to sleep.
If he sleeps, he sleeps continuously no less than A hours.
For some strange reason, the cat can not stay awake continuously more than B hours.
The cat is lazy, it could sleep all the time,
but sometimes interesting events occur(such as cat food, TV show, etc) .
The cat loves these events very much.
So, Please help the cat plan their day so as not to miss any interesting events.
Every day the cat wants to live on the same schedule.

Input The first line of the input file contains two integers A and B (1 <= A, B <= 24).
The second line of the input file contains the number n, the number of interesting events (1 <= n <= 20).
Following n rows describe the interesting events.
Each event is described line of the form hh:mm-hh:mm, which specifies
the time period during which it occurs. Time varies from 00:00 to 23:59.
No two interesting events will overlap.
If the event is completed earlier than the beginning, This means that it captures midnight.
The event is considered to be occupying the whole minute,
when it begins and the moment when it ends (event 12:00-13:59 lasted exactly 120 minutes). Start time and time end of the event are different.

Output If the cat can organize their day so that during all the interesting events not sleep, print to output file Yes.

On the second line Bring k - how many times a day cat should go to bed.
In the following k rows Bring out the intervals in which the cat sleeps in the same format, in which interesting events are set in the input file. If making a few, display any.

If the cat can not organize their day desired way, print to the output file No.

Sample Input

12 12
1
23:00-01:00
3 4
3
07:00-08:00
11:00-11:09
19:00-19:59

Sample Output

Yes
1
01:07-22:13
No
細節很多，比賽的時候卡了全場，實在是感人肺腑，不能多說#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define Maxn 100005
#define Maxm 1000005
#define Inf (1LL<<62)
#define inf 0x3f3f3f3f
#define eps 1e-8
#define lowbit(x) x&(-x)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define PI acos(-1.0)
#define make_pair MP
#define LL long long 
#define re freopen("in.txt","r",stdin)
#define wr freopen("out.txt","w",stdout)
using namespace std;
struct Node
{
    int st,ed,t;
    bool friend operator <(Node a,Node b)
    {
        return a.stb*60)
        {
            puts("No");
            continue;
        }
        int tt=1440-time[0].t;
        if(n==1)
        {
            if(tt>=a*60)
            {
                puts("Yes");
                puts("1");
                int hh1=((time[0].ed+1)/60)%24;
                int mm1=(time[0].ed+1)%60;
                int hh2=time[0].st-1>=0?((time[0].st-1)/60)%24:23;
                int mm2=((time[0].st-1)+60)%60;
                printf("%02d:%02d-%02d:%02d\n",hh1,mm1,hh2,mm2);
            }
            else
                puts("No");
            continue;
        }
        sort(time,time+n);
        tt=time[0].st+1440-time[n-1].ed-1;
        int tmp;
        if(tt>=a*60)
        	tmp=0;
        else
        {
        	tmp=tt+time[n-1].t;
        	int p=n-1;
        	while(p>=1&&time[p].st-time[p-1].ed-1b*60)
            {
                flag=false;
                break;
            }
            if(i!=n-1)
            {
            	if(time[i+1].st-time[i].ed-1>=a*60)
            	{
                	ans[++k].st=time[i].ed+1;
                	ans[k].ed=time[i+1].st-1;
                	tmp=0;
            	}
            	else 
                	tmp+=time[i+1].st-1-time[i].ed;   
            	if(tmp>b*60)
           	 	{
                	flag=false;
               	 	break;
            	}
            }        
        }
        if(tt>=a*60)
        {
            k++;
            ans[k].st=time[n-1].ed+1;ans[k].ed=time[0].st-1;
        }
        if(flag&&k)
        {
            puts("Yes");
            printf("%d\n",k);
            for(int i=1;i<=k;i++)
            {
                int hh1=(ans[i].st/60)%24;
                int mm1=(ans[i].st+60)%60;
                int hh2=ans[i].ed>=0?(ans[i].ed/60)%24:23;
                int mm2=(ans[i].ed+60)%60;
                printf("%02d:%02d-%02d:%02d\n",hh1,mm1,hh2,mm2);
            }
        }
        else
            puts("No");
    }
    return 0;
}