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: 3433
    User: HansBug
    Language: Pascal
    Result: Accepted
    Time:0 ms
    Memory:304 kb
****************************************************************/
 
var
   i,j,k,l,m,n,ans,l1,l2:longint;
   a:array[0..10000,1..2] 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 2,2];
               repeat
                     while a[i,2]<x do inc(i);
                     while a[j,2]>x do dec(j);
                     if i<=j then
                        begin
                             swap(a[i,1],a[j,1]);
                             swap(a[i,2],a[j,2]);
                             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:=1 to n do readln(a[i,1],a[i,2]);
     sort(1,n);
     for i:=1 to n do
         begin
              if (a[i,1]>=l1) then
                 begin
                      l1:=a[i,2];
                      inc(ans);
                 end
              else if (a[i,1]>=l2) then
                   begin
                        l2:=a[i,2];
                        inc(ans);
                   end;
              if l1<l2 then swap(l1,l2);
         end;
     writeln(ans);
     readln;
end.