习题2.2：A. Cleaning Shifts

Description

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.

理解：

给定一个时间T和N个时间区间，求最少需要多少个区间覆盖总区间[1,T]，无法覆盖区域[1,T]时输出-1。
类似于工作排序问题：
贪心策略：在符合时间情况的选项中选择结束时间最迟的牛
具体步骤：按照开始时间升序排列，如果开始时间相同，按照结束时间升序排列设t为最终结束时间。两层循环嵌套第一层是寻找每一个时间段开始工作的牛，第二层是寻找在同一时间段开始工作的牛中的最晚结束的牛。如果外层循环有一个点没有找到，则退出循环返回-1；

题解：

#include
#include
#include
using namespace std;
struct node
{
    int x,y;
    bool operator<(node t)const    //重载<运算符 
    {
        return x1)       //没有牛在第一个时段打扫 
    {
        puts("-1");
        return 0;
    }
    int p=0,ans=0;               
    for(int i=0; i=T) break;//这个语句也是很关键的
        }
        else break;  
    if(p