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BZOJ3314: [Usaco2013 Nov]Crowded Cows
3314: [Usaco2013 Nov]Crowded Cows
Time Limit: 1 Sec Memory Limit: 128 MBSubmit: 86 Solved: 61
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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
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
OUTPUT DETAILS: The cows at positions x=5 and x=6 are both crowded.
HINT
Source
Silver
1 #include<cstdio> 2 #include<cstdlib> 3 #include<cmath> 4 #include<cstring> 5 #include<algorithm> 6 #include<iostream> 7 #include<vector> 8 #include<map> 9 #include<set>10 #include<queue>11 #include<string>12 #define inf 100000000013 #define maxn 100000+100014 #define maxm 500+10015 #define eps 1e-1016 #define ll long long17 #define pa pair<int,int>18 #define for0(i,n) for(int i=0;i<=(n);i++)19 #define for1(i,n) for(int i=1;i<=(n);i++)20 #define for2(i,x,y) for(int i=(x);i<=(y);i++)21 #define for3(i,x,y) for(int i=(x);i>=(y);i--)22 #define mod 100000000723 using namespace std;24 inline int read()25 {26 int x=0,f=1;char ch=getchar();27 while(ch<‘0‘||ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();}28 while(ch>=‘0‘&&ch<=‘9‘){x=10*x+ch-‘0‘;ch=getchar();}29 return x*f;30 }31 struct rec{int x,y;}a[maxn],q[maxn];32 int n,m;33 bool can1[maxn],can2[maxn];34 inline bool cmp(rec a,rec b)35 {36 return a.x<b.x;37 }38 int main()39 {40 freopen("input.txt","r",stdin);41 freopen("output.txt","w",stdout);42 n=read();m=read();43 for1(i,n)a[i].x=read(),a[i].y=read();44 sort(a+1,a+n+1,cmp);45 int l=1,r=0;46 for1(i,n)47 {48 while(l<=r&&q[r].y<a[i].y)r--;49 q[++r]=a[i];50 while(l<=r&&q[l].x<a[i].x-m)l++;51 if(q[l].y>=a[i].y*2)can1[i]=1;52 }53 l=1,r=0;54 for3(i,n,1)55 {56 while(l<=r&&q[r].y<a[i].y)r--;57 q[++r]=a[i];58 while(l<=r&&q[l].x>a[i].x+m)l++;59 if(q[l].y>=a[i].y*2)can2[i]=1;60 }61 int ans=0;62 for1(i,n)if(can1[i]&&can2[i])ans++;63 printf("%d\n",ans);64 return 0;65 }
BZOJ3314: [Usaco2013 Nov]Crowded Cows