挤奶牛Crowded Cows 洛谷p3088
题目描述
Farmer John's N cows (1 <= N <= 50,000) are grazing along a one-dimensional fence. Cow i is standing at location x(i) and has height h(i) (1 <= x(i),h(i) <= 1,000,000,000).
A cow feels "crowded" if there is another cow at least twice her height within distance D on her left, and also another cow at least twice her height within distance D on her right (1 <= D <= 1,000,000,000). Since crowded cows produce less milk, Farmer John would like to count the number of such cows. Please help him.
FJ有N(1 <= N <= 50,000)头奶牛沿着一维的栅栏吃草，第i头奶牛在目标点x(i) ，它的身高是 h(i) (1 <=x(i),h(i) <= 1,000,000,000)。
当一头奶牛左边D距离内而且右边D距离内有身高至少是它的两倍的奶牛，t (1 <= D <= 1,000,000,000)，它就会觉得拥挤。
请计算觉得拥挤的奶牛的数量。
输入输出格式
输入格式：
* Line 1: Two integers, N and D.
* Lines 2..1+N: Line i+1 contains the integers x(i) and h(i). The locations of all N cows are distinct.
输出格式：
* Line 1: The number of crowded cows.
输入输出样例
输入样例#1： 复制
6 4 10 3 6 2 5 3 9 7 3 6 11 2
输出样例#1： 复制
2
说明
There are 6 cows, with a distance threshold of 4 for feeling crowded. Cow #1 lives at position x=10 and has height h=3, and so on.
The cows at positions x=5 and x=6 are both crowded.
#include
#define f(i,l,r) for(i=(l);i<=(r);i++)
using namespace std;
const int MAXN=50005;
int n,D;
int tree[MAXN<<2];
int ans;
struct Node{
int x,h,l,r;
bool operator < (const Node& tmp)const{
return x>1;
build(x<<1,l,mid);
build(x<<1|1,mid+1,r);
pushup(x);
}
inline void query(int x,int l,int r,int sj,int tj)
{
if(sj<=l&&r<=tj){
ans=max(ans,tree[x]);
return;
}
int mid=l+r>>1;
if(mid>=sj) query(x<<1,l,mid,sj,tj);
if(mid+1<=tj) query(x<<1|1,mid+1,r,sj,tj);
}
int main()
{
ios::sync_with_stdio(false);
int i,j,pos;
int res=0;
cin>>n>>D;
f(i,1,n){
cin>>a[i].x>>a[i].h;
}
sort(a+1,a+1+n);
f(i,1,n){
Node tmp;
tmp.x=a[i].x-D;
pos=lower_bound(a+1,a+1+n,tmp,cmp)-a;
a[i].l=pos;
tmp.x=a[i].x+D;
pos=lower_bound(a+1,a+1+n,tmp,cmp)-a;
if(a[pos].x>a[i].x+D) pos--;
a[i].r=pos;
}
build(1,1,n);
f(i,2,n-1){
ans=0;
query(1,1,n,a[i].l,i);
if(ans=a[i].h+a[i].h) res++;
}
cout<
#include
#define f(i,l,r) for(i=(l);i<=(r);i++)
#define ff(i,r,l) for(i=(r);i>=(l);i--)
using namespace std;
const int MAXN=50005;
struct Node{
int x,h,l=0,r=0;
bool operator < (const Node& tmp)const{
return x q;
int main()
{
ios::sync_with_stdio(false);
int i,j;
int ans=0;
cin>>n>>D;
f(i,1,n){
cin>>a[i].x>>a[i].h;
}
sort(a+1,a+1+n);
f(i,1,n){
while(!q.empty()&&a[q.front()].x+D=a[i].h+a[i].h) a[i].l=1;
while(!q.empty()&&a[q.back()].h<=a[i].h) q.pop_back();
q.push_back(i);
}
while(!q.empty()) q.pop_back();
ff(i,n,1){
while(!q.empty()&&a[q.front()].x>a[i].x+D) q.pop_front();
if(!q.empty()&&a[q.front()].h>=a[i].h+a[i].h) a[i].r=1;
while(!q.empty()&&a[q.back()].h<=a[i].h) q.pop_back();
q.push_back(i);
}
f(i,1,n){
if(a[i].l&&a[i].r) ans++;
}
cout<