3893: [Usaco2014 Dec]Cow Jog

3893: [Usaco2014 Dec]Cow Jog

Time Limit: 10 Sec  Memory Limit: 128 MB
Submit: 174  Solved: 87
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Description

The cows are out exercising their hooves again! There are N cows jogging on an infinitely-long single-lane track (1 <= N <= 100,000). Each cow starts at a distinct position on the track, and some cows jog at different speeds. With only one lane in the track, cows cannot pass each other. When a faster cow catches up to another cow, she has to slow down to avoid running into the other cow, becoming part of the same running group. The cows will run for T minutes (1 <= T <= 1,000,000,000). Please help Farmer John determine how many groups will be left at this time. Two cows should be considered part of the same group if they are at the same position at the end of T minutes.
在一条无限长的跑道上有N头牛，每头牛有自己的初始位置及奔跑的速度。牛之间不能互相穿透。当一只牛追上另一只牛时，它不得不慢下来，成为一个群体。求T分钟后一共有几个群体。

 

Input

The first line of input contains the two integers N and T. The following N lines each contain the initial position and speed of a single cow. Position is a nonnegative integer and speed is a positive integer; both numbers are at most 1 billion. All cows start at distinct positions, and these will be given in increasing order in the input.

 

Output

A single integer indicating how many groups remain after T minutes.

 

Sample Input

5 3
0 1
1 2
2 3
3 2
6 1

Sample Output

3

HINT

 

Source

Silver

 

题解：显然，一头牛的速度不会被后面的牛所影响，于是直接O(n)扫一遍，像打擂台一样即可，这样子分堆数也就出来啦

/**************************************************************
    Problem: 3893
    User: HansBug
    Language: Pascal
    Result: Accepted
    Time:1204 ms
    Memory:1008 kb
****************************************************************/
 
var
   i,j,k,l,m,n,ans:longint;
   a,b:array[0..100100] of longint;
   t:int64;
begin
     readln(n,t);
     for i:=1 to n do readln(a[i],b[i]);
     a[n+1]:=maxlongint;
     b[n+1]:=maxlongint;
     l:=n+1;
     for i:=n downto 1 do
         begin
              if (a[i]+(b[i]*t))<(a[l]+(b[l]*t)) then
                 begin
                      inc(ans);
                      l:=i;
                 end;
         end;
     writeln(ans);
end.