1664: [Usaco2006 Open]County Fair Events 参加节日庆祝

1664: [Usaco2006 Open]County Fair Events 参加节日庆祝

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 255  Solved: 185
[Submit][Status][Discuss]

Description

Farmer John has returned to the County Fair so he can attend the special events (concerts, rodeos, cooking shows, etc.). He wants to attend as many of the N (1 <= N <= 10,000) special events as he possibly can. He's rented a bicycle so he can speed from one event to the next in absolutely no time at all (0 time units to go from one event to the next!). Given a list of the events that FJ might wish to attend, with their start times (1 <= T <= 100,000) and their durations (1 <= L <= 100,000), determine the maximum number of events that FJ can attend. FJ never leaves an event early.

有N个节日每个节日有个开始时间,及持续时间. 牛想尽可能多的参加节日,问最多可以参加多少. 注意牛的转移速度是极快的,不花时间.

Input

* Line 1: A single integer, N.

* Lines 2..N+1: Each line contains two space-separated integers, T and L, that describe an event that FJ might attend.

Output

* Line 1: A single integer that is the maximum number of events FJ can attend.

Sample Input

7
1 6
8 6
14 5
19 2
1 8
18 3
10 6

INPUT DETAILS:

Graphic picture of the schedule:
11111111112
12345678901234567890---------这个是时间轴.
--------------------
111111 2222223333344
55555555 777777 666

这个图中1代表第一个节日从1开始,持续6个时间,直到6.

Sample Output

4

OUTPUT DETAILS:

FJ can do no better than to attend events 1, 2, 3, and 4.

HINT

 

Source

Silver

题解：一个考得烂了的贪心，很经典的最多不向交区间问题而已

 /**************************************************************
     Problem:
     User: HansBug
     Language: Pascal
     Result: Accepted
     Time: ms
     Memory: kb
 ****************************************************************/
 
 var
    i,j,k,l,m,n: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 ,];y:=a[(l+r) div ,];
                repeat
                      while (a[i,]<x) or ((a[i,]=x) and (a[i,]>y)) do inc(i);
                      while (a[j,]>x) or ((a[j,]=x) and (a[j,]<y)) 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,]);
      for i:= to n do a[i,]:=a[i,]+a[i,]-;
      sort(,n);l:=;
      for i:= to n do if a[i,]<>a[l,] then
          begin
               inc(l);
               a[l,]:=a[i,];
               a[l,]:=a[i,];
          end;
      n:=l;l:=;k:=;
      for i:= to n do
          if a[i,]>l then
             begin
                  inc(k);
                  l:=a[i,];
             end;
      writeln(k);
      readln;
 end.