poj 2376 Cleaning Shifts

http://poj.org/problem?id=2376

Cleaning Shifts

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12604		Accepted: 3263

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.

 

 

分析：

題意就是給你N段小區間 [n , m]和一個T，T表明大區間[1 ，T] ， 要求你找出最少使用小區間徹底覆蓋大區間。

一開始一直WA，由於理解錯了「徹底覆蓋」的概念。便是下面的數據：

1 　　　　3 10　　　　　　//3表明小區間個數，10表明大區間長度。
2 　　　　1 5
3 　　　　6 8
4 　　　　9 10

上面的數據個人程序輸出的是：-1

由於有兩個區間沒有「徹底覆蓋」，[5 ，6]和[8 ，9]。

可是正確答案是 3 。

WA代碼：

 1 #include<cstdio>
 2 #include<algorithm>
 3 using namespace std;
 4 struct P
 5 {
 6     int x,y;
 7     bool operator < (const P & p) const
 8     {
 9         return x < p.x || (x == p.x && y > p.y);
10     }
11 }a[25010];
12 
13 int main()
14 {
15     int n,t;
16     while(~scanf("%d %d",&n,&t))
17     {
18         for(int i = 0;i < n;i++) 
19             scanf("%d %d",&a[i].x,&a[i].y);
20         sort(a ,a + n);
21         int res = 1,s;
22         if(a[0].x > 1)
23             printf("-1\n");
24         else
25         {
26             s = a[0].y;
27             for(int i = 1;i < n && s < t;)
28             {
29                 int tmp = 0;
30                 while(i < n && a[i].x <= s)
31                 {
32                     tmp = max(tmp , a[i].y);
33                     i++;
34                 }
35                 if(tmp > s)
36                 {
37                     s = tmp;
38                     res++;
39                 }
40                 else
41                     break;
42             }
43         }
44         if(s >= t)
45             printf("%d\n",res);
46         else
47             printf("-1\n");
48     }
49     return 0;
50 }

View Code

 

AC代碼：

 1 #include<cstdio>
 2 #include<algorithm>
 3 using namespace std;
 4 
 5 struct P{
 6     int x,y;
 7     bool operator < (const P & p) const {
 8         return x < p.x || (x == p.x && y > p.y);
 9     }
10 }a[25010];
11 
12 int main() {
13     int n,t;
14     while(~scanf("%d %d",&n,&t)) {
15         for(int i = 0;i < n;i++) 
16             scanf("%d %d",&a[i].x,&a[i].y);
17             
18         sort(a ,a + n);
19         
20         int res = 1,s;
21         if(a[0].x > 1) {
22             printf("-1\n");
23             continue;
24         } else {
25             s = a[0].y;
26             for(int i = 1;i < n && s < t;) {
27                 int tmp = 0;
28                 while(i < n && a[i].x <= s + 1) {
29                     tmp = max(tmp , a[i].y);
30                     i++;
31                 }
32                 if(tmp > s) {
33                     s = tmp;
34                     res++;
35                 } else break;
36             }
37         }
38             
39         if(s >= t) {
40             printf("%d\n",res);
41         } else {
42             printf("-1\n");
43         }
44     }
45     return 0;
46 }