题目地址：
http://poj.org/problem?id=2376
题目内容：
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 12226 | Accepted: 3187 |
Description
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
* 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
Sample Input
3 10
1 7
3 6
6 10
Sample Output
2
Hint
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
#include <algorithm>
#include <cstdio>
using namespace std; int n,t;
struct cow
{
int start;
int fin;
};
cow whole[]; bool cmp(const cow& a, const cow& b)
{
return a.start < b.start;
} int main(void)
{
scanf("%d%d", &n, &t);
for (int i = ; i < n; i ++) {
int t1,t2;
scanf("%d%d", &t1, &t2);
whole[i].start = t1;
whole[i].fin = t2;
}
sort(whole, whole + n, cmp);
int top = ;
int max_length = ;
int res = ;
for (int i = ; i < n; i ++) { if (whole[i].start > top) {
if (max_length == ) {
res = -;
break;
} else {
res ++;
top = max_length + ;
max_length = ;
if (top > t)
break;
}
}
if (whole[i].start <= top && whole[i].fin >= top && max_length < whole[i].fin) {
max_length = whole[i].fin;
}
//printf("now is %d, top is %d, max is %d\n", i, top, max_length);
}
if (max_length != ) {
res ++;
top = max_length + ;
}
if (top <= t)
res = -;
printf("%d\n", res);
}