[USACO13NOV]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 #1

6 4
10 3
6 2
5 3
9 7
3 6
11 2

输出 #1

2

跑两遍单调队列，每一遍都维护一个单调递减的队列，对符合的元素进行标记；当一个元素被标记两遍以后，符合要求；两遍单调队列难想，其他的都是单调队列板子；

代码：

#include<bits/stdc++.h>
using namespace std;
struct Node{int x,h;
}a[50100];
bool cmp(Node p,Node q){return p.x<q.x;
}
struct Nod{int post,date;
}sta[50100];
bool vis[50100][3];
int main(){int n,d;scanf("%d%d",&n,&d);for(int i=1;i<=n;i++){scanf("%d%d",&a[i].x,&a[i].h);}sort(a+1,a+n+1,cmp);int ans=1,head=1;for(int i=1;i<=n;i++){while(head<=ans&&sta[ans].date<=a[i].h){ans--;}sta[++ans].date=a[i].h;sta[ans].post=a[i].x;if(ans==1) continue;while(head<=ans&&sta[head].post<a[i].x-d){head++;}if(sta[head].date>=2*a[i].h){//cout<<i<<" "<<a[i].x<<endl; vis[i][1]=true;}}for(int i=1;i<=n;i++){sta[i].date=sta[i].post=0;//清0，特别注意}ans=1,head=1;for(int i=n;i>=1;i--){while(head<=ans&&sta[ans].date<=a[i].h){ans--;}sta[++ans].date=a[i].h;sta[ans].post=a[i].x;if(ans==1) continue;while(head<=ans&&sta[head].post>a[i].x+d){head++;}if(sta[head].date>=2*a[i].h){//cout<<i<<" "<<a[i].x<<endl;vis[i][2]=true;}}int sum=0;for(int i=1;i<=n;i++){if(vis[i][1]&&vis[i][2]){sum++;}}printf("%d\n",sum);return 0;
}

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