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
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
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
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.