BZOJ3314: [Usaco2013 Nov]Crowded Cows

3314: [Usaco2013 Nov]Crowded Cows

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 86  Solved: 61
[Submit][Status]

Description

 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.

N头牛在一个坐标轴上，每头牛有个高度。现给出一个距离值D。

如果某头牛在它的左边，在距离D的范围内，如果找到某个牛的高度至少是它的两倍，且在右边也能找到这样的牛的话。则此牛会感觉到不舒服。

问有多少头会感到不舒服。

Input

* 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.

Output

* Line 1: The number of crowded cows.

Sample Input

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

INPUT DETAILS: 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.

Sample Output

2
OUTPUT DETAILS: The cows at positions x=5 and x=6 are both crowded.

HINT

 

Source

Silver

题解：

单调队列维护可行域内的最大值。

1A是对我的一种补偿吗？T_T

代码：

#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<string>
#define inf 1000000000
#define maxn 100000+1000
#define maxm 500+100
#define eps 1e-10
#define ll long long
#define pa pair<int,int>
#define for0(i,n) for(int i=0;i<=(n);i++)
#define for1(i,n) for(int i=1;i<=(n);i++)
#define for2(i,x,y) for(int i=(x);i<=(y);i++)
#define for3(i,x,y) for(int i=(x);i>=(y);i--)
#define mod 1000000007
using namespace std;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();}
    return x*f;
}
struct rec{int x,y;}a[maxn],q[maxn];
int n,m;
bool can1[maxn],can2[maxn];
inline bool cmp(rec a,rec b)
{
    return a.x<b.x;
}
int main()
{
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    n=read();m=read();
    for1(i,n)a[i].x=read(),a[i].y=read();
    sort(a+1,a+n+1,cmp);
    int l=1,r=0;
    for1(i,n)
    {
        while(l<=r&&q[r].y<a[i].y)r--;
        q[++r]=a[i];
        while(l<=r&&q[l].x<a[i].x-m)l++;
        if(q[l].y>=a[i].y*2)can1[i]=1;
    }
    l=1,r=0;
    for3(i,n,1)
    {
        while(l<=r&&q[r].y<a[i].y)r--;
        q[++r]=a[i];
        while(l<=r&&q[l].x>a[i].x+m)l++;
        if(q[l].y>=a[i].y*2)can2[i]=1;
    }
    int ans=0;
    for1(i,n)if(can1[i]&&can2[i])ans++;
    printf("%d\n",ans);
    return 0;
}