Cleaning Shifts POJ - 2376（贪心）

Farmer John is assigning some of his N (1 <= N <= 25,000) cows to do some cleaning chores around the barn. He always wants to have one cow working on cleaning things up and has divided the day into T shifts (1 <= T <= 1,000,000), the first being shift 1 and the last being shift T.

Each cow is only available at some interval of times during the day for work on cleaning. Any cow that is selected for cleaning duty will work for the entirety of her interval.

Your job is to help Farmer John assign some cows to shifts so that (i) every shift has at least one cow assigned to it, and (ii) as few cows as possible are involved in cleaning. If it is not possible to assign a cow to each shift, print -1.
Input
* Line 1: Two space-separated integers: N and T

• Lines 2..N+1: Each line contains the start and end times of the interval during which a cow can work. A cow starts work at the start time and finishes after the end time.
  Output
• Line 1: The minimum number of cows Farmer John needs to hire or -1 if it is not possible to assign a cow to each shift.
  Sample Input
  3 10
  1 7
  3 6
  6 10
  Sample Output
  2
  Hint
  This problem has huge input data,use scanf() instead of cin to read data to avoid time limit exceed.

INPUT DETAILS:

There are 3 cows and 10 shifts. Cow #1 can work shifts 1..7, cow #2 can work shifts 3..6, and cow #3 can work shifts 6..10.

OUTPUT DETAILS:

By selecting cows #1 and #3, all shifts are covered. There is no way to cover all the shifts using fewer than 2 cows.

思路：这道题是典型的贪心，但是在实现上有有些细节都要非常注意才行。
先排序，然后一个一个考虑。
其实觉得我写的还不够简洁，还可以简化一些。
代码：

#include<iostream>
#include<cstdio>
#include<stack>
#include<algorithm>
#include<queue>
#include<cstring>
#include<cmath>
#define maxx 100005
using namespace std;
struct node
{
    int l,r;
}p[25005];
bool cmp(node x1,node x2)
{
    if(x1.l==x2.l)
        return x1.r>x2.r;
    return x1.l<x2.l;
}
int main()
{
    int n,m;
    cin>>n>>m;
    for(int i=0;i<n;i++)
    {
        scanf("%d%d",&p[i].l,&p[i].r);
        if(p[i].r>m)
            p[i].r=m;
    }
    sort(p,p+n,cmp);
    if(p[0].l>1)
    {
        cout<<-1<<endl;
        return 0;
    }
    int curIndex;
    int cover=p[0].r;
    bool sign=false;
    int i=1;
    int countt=1;
    while(i<n)
    {
        if(sign)
        {
            if(p[i].r<=p[curIndex].r)
            {
                i++;
                continue;
            }
            if(cover+1<p[i].l)
            {
                cover=p[curIndex].r;
                countt++;
                if(cover+1<p[i].l)
                {
                    cout<<-1<<endl;
                    return 0;
                }
                curIndex=i;
            }
            else
            {
                if(p[i].r>p[curIndex].r)
                    curIndex=i;
            }
        }
        else
        {
            if(p[i].r<=cover)
            {
                i++;
                continue;
            }
            if(cover+1<p[i].l)
                {
                    cout<<-1<<endl;
                    return 0;
                }
            curIndex=i;
            sign=true;
        }
        i++;
    }
    if(!sign)
    {
        if(cover==m)
            cout<<countt<<endl;
        else
            cout<<-1<<endl;
    }
    else
    {
        countt++;
        if(p[curIndex].r==m)
            cout<<countt<<endl;
        else
            cout<<-1<<endl;
    }
    return 0;
}