5163: 第k大斜率
Time Limit: 10 Sec Memory Limit: 512 MBSubmit: 15 Solved: 4
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Description
Input
Output
输出一行,包含一个整数,表示第k小的斜率向下取整的结果。
Sample Input
-1 -1
2 1
3 3
1 4
Sample Output
#include<bits/stdc++.h> #define ll long long #define maxn 100005 using namespace std; int l,r,mid,ans; const int inf=1000000000; int px[maxn],ky; struct node{ int x,y; ll tmp; bool operator <(const node& u)const{ return tmp==u.tmp?x<u.x:tmp<u.tmp; } }a[maxn]; int n,m,f[maxn]; ll k; inline int query(int x){ int an=0; for(;x;x-=x&-x) an+=f[x]; return an; } inline void update(int x,int y){ for(;x<=ky;x+=x&-x) f[x]+=y; } inline ll calc(){ ll an=0; memset(f,0,sizeof(f)); for(int i=1;i<=n;i++) a[i].tmp=(ll)a[i].y-mid*(ll)px[a[i].x]; sort(a+1,a+n+1); for(int i=1;i<=n;i++){ an+=(ll)query(a[i].x-1); update(a[i].x,1); } return an; } int main(){ scanf("%d%lld",&n,&k); for(int i=1;i<=n;i++){ scanf("%d%d",&a[i].x,&a[i].y); px[i]=a[i].x; } sort(px+1,px+n+1); ky=unique(px+1,px+n+1)-px-1; for(int i=1;i<=n;i++) a[i].x=lower_bound(px+1,px+ky+1,a[i].x)-px; l=-inf,r=inf; while(l<=r){ mid=l+r>>1; if(calc()>=k) ans=mid,l=mid+1; else r=mid-1; } printf("%d\n",ans); return 0; }