Stall Reservations
Time Limit: 1000MS | | Memory Limit: 65536K | ||
Total Submissions: 4274 | | Accepted: 1530 | | Special Judge |
Description
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
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.
<span style="font-size:18px;color:#3366ff;">#include <iostream>
#include<cstdio>
#include<queue>
#include<algorithm>
#include<cstring>
using namespace std;
struct Node{
int s,e,id,num;
bool operator<(const Node &a) const
{
return a.e<e;
}
}node[50005];
int cmp(Node a,Node b)
{
if(a.s!=a.s)
return a.e<b.e;
else return a.s<b.s;
}
int cmp2(Node a,Node b)
{
return a.id<b.id;
}
int main()
{
int n;
while(~scanf("%d",&n))
{
for(int i=1;i<=n;i++)
{
scanf("%d %d",&node[i].s,&node[i].e);
node[i].id=i;
}
sort(node+1,node+n+1,cmp);
priority_queue<Node> q;
node[1].num=1;
q.push(node[1]);
int cnt=1;
for(int i=2;i<=n;i++)
{
Node temp=q.top();
if(node[i].s<=temp.e)
{
cnt++;
node[i].num=cnt;
q.push(node[i]);
}
else
{
node[i].num=temp.num;
q.pop();
q.push(node[i]);
}
}
sort(node+1,node+n+1,cmp2);
printf("%d\n",cnt);
for(int i=1;i<=n;i++)
printf("%d\n",node[i].num);
}
return 0;
}</span>
