3433: [Usaco2014 Jan]Recording the Moolympics

3433: [Usaco2014 Jan]Recording the Moolympics

Time Limit: 10 Sec  Memory Limit: 128 MB
Submit: 137  Solved: 89
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Description

Being a fan of all cold-weather sports (especially those involving cows), Farmer John wants to record as much of the upcoming winter Moolympics as possible. The television schedule for the Moolympics consists of N different programs (1 <= N <= 150), each with a designated starting time and ending time. FJ has a dual-tuner recorder that can record two programs simultaneously. Please help him determine the maximum number of programs he can record in total.

给出n个区间[a,b).有2个记录器.每个记录器中存放的区间不能重叠.

求2个记录器中最多可放多少个区间.

Input

* Line 1: The integer N.

* Lines 2..1+N: Each line contains the start and end time of a single program (integers in the range 0..1,000,000,000).

Output

* Line 1: The maximum number of programs FJ can record.

Sample Input

6
0 3
6 7
3 10
1 5
2 8
1 9

INPUT DETAILS: The Moolympics broadcast consists of 6 programs. The first runs from time 0 to time 3, and so on.

Sample Output

4

OUTPUT DETAILS: FJ can record at most 4 programs. For example, he can record programs 1 and 3 back-to-back on the first tuner, and programs 2 and 4 on the second tuner.

HINT

 

Source

Silver 鸣谢Alegria_提供译文

题解：（呵呵哒我会告诉你我的第一反应是网络流？）

其实仔细看看后发现还是个贪心，只不过现在是两个容器= =，然后就是经典的右边界排序后O(n)乱搞了

 /**************************************************************
     Problem:
     User: HansBug
     Language: Pascal
     Result: Accepted
     Time: ms
     Memory: kb
 ****************************************************************/

 var
    i,j,k,l,m,n,ans,l1,l2:longint;
    a:array[..,..] of longint;
 procedure swap(var x,y:longint);
           var z:longint;
           begin
                z:=x;x:=y;y:=z;
           end;
 procedure sort(l,r:longint);
           var i,j,x,y:longint;
           begin
                i:=l;j:=r;x:=a[(l+r) div ,];
                repeat
                      while a[i,]<x do inc(i);
                      while a[j,]>x do dec(j);
                      if i<=j then
                         begin
                              swap(a[i,],a[j,]);
                              swap(a[i,],a[j,]);
                              inc(i);dec(j);
                         end;
                until i>j;
                if i<r then sort(i,r);
                if l<j then sort(l,j);
           end;

 begin
      readln(n);
      for i:= to n do readln(a[i,],a[i,]);
      sort(,n);
      for i:= to n do
          begin
               if (a[i,]>=l1) then
                  begin
                       l1:=a[i,];
                       inc(ans);
                  end
               else if (a[i,]>=l2) then
                    begin
                         l2:=a[i,];
                         inc(ans);
                    end;
               if l1<l2 then swap(l1,l2);
          end;
      writeln(ans);
      readln;
 end.