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
题目：
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。
例如T=10，有3个区间[1,7],[3,6],[8,10],则最少需要两个区间来覆盖，选择区间1和区间3。
解题思路：
使用贪心法。首先将区间按开始时间从小到大排序，开始时间相等按结束时间从小到大排序。
1 如果第一个区间不是从1开始，则无法覆盖，输出-1。
2 令当前覆盖到的时间time为开始时间为1的区间中结束时间的最大值。
3 从开始时间不为1的区间开始循环遍历。
4 选择合法区间中结束时间值最大的那个区间，合并到[1,time]，合并后time为选择的区间的结束时间。所谓合法区间是指区间的起始时间start<=time+1，这样才能和区间[1,time]合并。如果没有找到合法区间或者找到的区间结束时间比time小，则无法覆盖，结束循环。
5 循环结束后，根据time是否大于等于T，输出结果。
1 #include<iostream>
2 #include<cstdio>
3 #include<cstring>
4 #include<algorithm>
5 #include<stdlib.h>
6 using namespace std;
7 #define N 26000
8 struct Node
9 {
10 int st,ed;
11 }p[N];
12 bool cmp(Node a,Node b){
13 if(a.st!=b.st)
14 return a.st<b.st;
15 return a.ed<b.ed;
16 }
17 int main()
18 {
19 int n,t;
20 while(scanf("%d%d",&n,&t)==2)
21 {
22 for(int i=0;i<n;i++)
23 {
24 scanf("%d%d",&p[i].st,&p[i].ed);
25 }
26
27 sort(p,p+n,cmp);
28
29 int i=0;
30 int time=p[i].ed;
31 if(p[i].st!=1)
32 {
33 printf("-1\n");
34 continue;
35 }
36 i++;
37 while(i<n && p[i].st==1)
38 {
39 time=p[i].ed;
40 i++;
41 }
42 int ans=1;
43
44 while(time<t)
45 {
46
47 if(i>=n)
48 break;
49 int tmp=i;
50 int w=p[i].ed;
51 i++;
52 while(i<n && p[i].st<=time+1)
53 {
54 if(w<p[i].ed)
55 {
56 tmp=i;
57 w=p[i].ed;
58 }
59 i++;
60 }
61
62 if(w<=time || p[tmp].st>time+1)
63 {
64 break;
65 }
66 else
67 {
68 time=w;
69 ans++;
70 }
71
72 }
73 if(time<t)
74 ans=-1;
75
76 printf("%d\n",ans);
77 }
78 return 0;
79 }