http://poj.org/problem?id=2376
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 12604 | Accepted: 3263 |
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.
10 //3代表小区间个数，10代表大区间长度。
上面的数据我的程序输出的是：-1
因为有两个区间没有“完全覆盖”，[5 ，6]和[8 ，9]。
但是正确答案是 3 。
WA代码：
#include<cstdio>
#include<algorithm>
using namespace std;
struct P
{
int x,y;
bool operator < (const P & p) const
{
return x < p.x || (x == p.x && y > p.y);
}
}a[]; int main()
{
int n,t;
while(~scanf("%d %d",&n,&t))
{
for(int i = ;i < n;i++)
scanf("%d %d",&a[i].x,&a[i].y);
sort(a ,a + n);
int res = ,s;
if(a[].x > )
printf("-1\n");
else
{
s = a[].y;
for(int i = ;i < n && s < t;)
{
int tmp = ;
while(i < n && a[i].x <= s)
{
tmp = max(tmp , a[i].y);
i++;
}
if(tmp > s)
{
s = tmp;
res++;
}
else
break;
}
}
if(s >= t)
printf("%d\n",res);
else
printf("-1\n");
}
return ;
}
AC代码：
#include<cstdio>
#include<algorithm>
using namespace std; struct P{
int x,y;
bool operator < (const P & p) const {
return x < p.x || (x == p.x && y > p.y);
}
}a[]; int main() {
int n,t;
while(~scanf("%d %d",&n,&t)) {
for(int i = ;i < n;i++)
scanf("%d %d",&a[i].x,&a[i].y); sort(a ,a + n); int res = ,s;
if(a[].x > ) {
printf("-1\n");
continue;
} else {
s = a[].y;
for(int i = ;i < n && s < t;) {
int tmp = ;
while(i < n && a[i].x <= s + ) {
tmp = max(tmp , a[i].y);
i++;
}
if(tmp > s) {
s = tmp;
res++;
} else break;
}
} if(s >= t) {
printf("%d\n",res);
} else {
printf("-1\n");
}
}
return ;
}