Stall Reservations
Oh those picky N (1 <= N <= 50,000) cows! They are so picky that each one will only be milked over some precise time interval A..B (1 <= A <= B <= 1,000,000), which includes both times A and B. Obviously, FJ must create a reservation system to determine which stall each cow can be assigned for her milking time. Of course, no cow will share such a private moment with other cows.
Help FJ by determining:
- The minimum number of stalls required in the barn so that each cow can have her private milking period
- An assignment of cows to these stalls over time
Many answers are correct for each test dataset; a program will grade your answer.
Input
Line 1: A single integer, N
Lines 2..N+1: Line i+1 describes cow i's milking interval with two space-separated integers.
Output
Line 1: The minimum number of stalls the barn must have.
Lines 2..N+1: Line i+1 describes the stall to which cow i will be assigned for her milking period.
Sample Input
5
1 10
2 4
3 6
5 8
4 7
Sample Output
4
1
2
3
2
4
Hint
Explanation of the sample:
Here's a graphical schedule for this output:
Time 1 2 3 4 5 6 7 8 9 10
Stall 1 c1>>>>>>>>>>>>>>>>>>>>>>>>>>>
Stall 2 .. c2>>>>>> c4>>>>>>>>> .. ..
Stall 3 .. .. c3>>>>>>>>> .. .. .. ..
Stall 4 .. .. .. c5>>>>>>>>> .. .. ..
Other outputs using the same number of stalls are possible. //！！！
首先根据挤奶时间的先后顺序排序。。。然后将第一头牛加入优先队列。。然后就是加入优先队列的牛应该根据越早结束挤奶那么优先级更高，如果时间结束点相等，那么开始时间早的优先级高。。。
然后从前向后枚举。如果碰到有牛的挤奶时间的开始值大于优先队列的首部的结束值，那么说明这两头牛可以一起公用一个挤奶房。。然后从优先队列中删除这头牛。。那么这个问题就得到解决了。。。
开始时间升序排序，从左往右排，不会出现排在队首左边的情况。（贪心）
总结：开始时间升序排，每个条件都用上，求含不重叠子序列的最少序列数；
结束时间升序排，只用部分条件，求一个序列含不重叠子序列最多数。
#include<stdio.h>
#include<algorithm>
#include<queue>
using namespace std; struct Node{
int x,y,no;
friend bool operator<(Node a,Node b)
{
if(a.y==b.y) return a.x>b.x;
return a.y>b.y;
}
}node[];
bool cmp(Node a,Node b)
{
if(a.x==b.x) return a.y<b.y;
return a.x<b.x;
}
int a[];
priority_queue<Node> q;
int main()
{
int n,c,i;
scanf("%d",&n);
for(i=;i<=n;i++){
scanf("%d%d",&node[i].x,&node[i].y);
node[i].no=i;
}
sort(node+,node+n+,cmp);
q.push(node[]);
a[node[].no]=;
c=;
for(i=;i<=n;i++){
if(q.size()&&node[i].x>q.top().y){
a[node[i].no]=a[q.top().no];
q.pop();
}
else{
c++;
a[node[i].no]=c;
}
q.push(node[i]);
}
printf("%d\n",c);
for(i=;i<=n;i++){
printf("%d\n",a[i]);
}
return ;
}