题目描述： John 有N 头牛，每头牛有自己工作时间段，开始时间和结束时间，对于给定的一个时间T，要保证每个时间段都有牛工作，值得注意的是若T=10，1-7与8-10可以算作一个解，求用最少牛的时候牛的数量。
http://poj.org/problem?id=2376 Cleaning Shifts Time
Limit: 1000MS Memory Limit: 65536K
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.
Source USACO 2004 December Silver
AC代码：
#include <iostream> #include <cstdio> #include <algorithm> #include <cstdlib> using namespace std;; struct stu { int l; int r; }a[25010]; bool cmp(stu a, stu b) { if(a.l == b.l) return a.r < b.r; else return a.l<b.l; } int main() { int N, T,i,x,y,m=0,t; scanf("%d%d%*c", &N, &T); for (i = 0; i < N; i++) scanf("%d%d%*c", &a[i].l, &a[i].r); int top = 0; x=0; sort(a,a+N,cmp); while(x<T) { int x1=x+1; for (i = top; i < N; i++) if(a[i].l <= x1 && a[i].r >= x1) x=x>a[i].r? x:a[i].r; else if (a[i].l > x1) {top = i; break;} if(x < x1) break; m++; } if(x<T) cout<<"-1"<<endl; else cout<<m<<endl; return 0; }解题思路： 贪心区间问题 1.将各个节点按开始的时间排序，当开始时间一样时，后结束的节点放前面。 2.以第一个节点为开始，向后寻找开始时间不超过第一个个节点的（结束时间+1）的节点集合中，结束时间最长的一个节点 。更新开始节点。 3.用top储存遍历开始的节点 。 4.m++记录用的牛的数目。
