1651: [Usaco2006 Feb]Stall Reservations 专用牛棚

Time Limit: 10 Sec Memory Limit: 64 MB
Submit: 566 Solved: 314
[Submit][Status]

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

有N头牛,每头牛有个喝水时间,这段时间它将专用一个Stall 现在给出每头牛的喝水时间段,问至少要多少个Stall才能满足它们的要求

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

OUTPUT DETAILS:

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.

HINT

不妨试下这个数据,对于按结束点SORT,再GREEDY的做法 1 3 5 7 6 9 10 11 8 12 4 13 正确的输出应该是3

Source

Silver

题解：呵呵呵呵呵，逗比的我一开始看到N<=50000，然后就开了100000的数组，然后欢乐满满地RE（TuT），然后加了一个0，AC。。。好啦，这个题就是统计一下同一时间最多有多少只牛在同时进行喝水即可，只要想办法用线性的时间做到维护当前值即可。。。

 1 var
 2    i,j,k,l,m,n:longint;
 3    a:array[0..1050000] of longint;
 4 begin
 5      readln(n);
 6      fillchar(a,sizeof(a),0);
 7      for i:=1 to n do
 8          begin
 9               readln(j,k);
10               inc(a[j]);
11               dec(a[k+1]);
12          end;
13      j:=0;l:=0;
14      for i:=1 to 1000000 do
15          begin
16               j:=j+a[i];
17               if j>l then l:=j;
18          end;
19      writeln(l);
20 end.
21