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=,f=;char ch=getchar();

     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

     while(ch>=''&&ch<=''){x=*x+ch-'';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+,a+n+,cmp);

     int l=,r=;

     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*)can1[i]=;

     }

     l=,r=;

     for3(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*)can2[i]=;

     }

     int ans=;

     for1(i,n)if(can1[i]&&can2[i])ans++;

     printf("%d\n",ans);

     return ;

 }

秒客网