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。

Silver

 1 /*by SilverN*/ 2 #include<algorithm> 3 #include<iostream> 4 #include<cstring> 5 #include<cstdio> 6 #include<cmath> 7 #define LL long long 8 using namespace std; 9 const int mxn=100010;10 LL read(){11     LL x=0,f=1;char ch=getchar();12     while(ch<‘0‘ || ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}13     while(ch>=‘0‘ && ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();}14     return x*f;15 }16 LL v[mxn],p[mxn];17 LL id[mxn];18 int n;LL T;19 int cmp(int a,int b){20     if(p[a]==p[b])return a>b;21     return p[a]<p[b];22 }23 int main(){24     n=(LL)read();25     T=read();26     int i,j;27     for(i=1;i<=n;i++){28         p[i]=read();v[i]=read();29         p[i]+=v[i]*T;30     }31     for(i=1;i<=n;i++)id[i]=i;32     sort(id+1,id+n+1,cmp);33     int last=0,ans=0;34     for(i=1;i<=n;i++){35         if(id[i]>last){ans++;last=id[i];}36     }37     printf("%d\n",ans);38     return 0;39 }