3314: [Usaco2013 Nov]Crowded Cows
Time Limit: 1 Sec Memory Limit: 128 MB
Submit: 111 Solved: 79
[Submit][Status][Discuss]
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
题解：一个萌萌哒RMQ问题（HansBug：吐槽下翻译——输入里面明明说了the location is distinct，也就是说每只牛坐标唯一，这样子虽然没有实质性变化，但是简化了不少问题，毕竟题目说是左边和右边的牛，没说此位置上的= =），将坐标排序，然后就是区间查询最大值啦，然后没了
1 /************************************************************** 2 Problem: 3314 3 User: HansBug 4 Language: Pascal 5 Result: Accepted 6 Time:608 ms 7 Memory:6008 kb 8 ****************************************************************/ 9 10 var 11 i,j,k,l,m,n:longint; 12 a,b:array[0..100000,1..2] of longint; 13 c:array[0..17,0..60000] of longint; 14 function max(x,y:longint):longint; 15 begin 16 if x>y then max:=x else max:=y; 17 end; 18 procedure swap(var x,y:longint); 19 var z:longint; 20 begin 21 z:=x;x:=y;y:=z; 22 end; 23 procedure sort(l,r:longint); 24 var i,j,x:longint; 25 begin 26 i:=l;j:=r;x:=a[(l+r) div 2,1]; 27 repeat 28 while a[i,1]<x do inc(i); 29 while a[j,1]>x do dec(j); 30 if i<=j then 31 begin 32 swap(a[i,1],a[j,1]); 33 swap(a[i,2],a[j,2]); 34 inc(i);dec(j); 35 end; 36 until i>j; 37 if i<r then sort(i,r); 38 if l<j then sort(l,j); 39 end; 40 function getmax(x,y:longint):longint; 41 var i:longint; 42 begin 43 if y<x then exit(-1); 44 i:=trunc(ln(y-x+1)/ln(2)); 45 exit(max(c[i,x],c[i,y-trunc(exp(ln(2)*i))+1])); 46 end; 47 begin 48 readln(n,m); 49 for i:=1 to n do readln(a[i,1],a[i,2]); 50 sort(1,n); 51 for i:=1 to n do c[0,i]:=a[i,2]; 52 for i:=1 to trunc(ln(n)/ln(2)+1) do 53 for j:=1 to n-trunc(exp(ln(2)*i))+1 do 54 c[i,j]:=max(c[i-1,j],c[i-1,j+trunc(exp(ln(2)*(i-1)))]); 55 k:=1;fillchar(b[1],sizeof(b[1]),0); 56 for i:=2 to n do 57 begin 58 while (a[k,1]+m)<a[i,1] do inc(k); 59 if getmax(k,i-1)>=(a[i,2]*2) then b[i,1]:=1; 60 end; 61 k:=n;fillchar(b[1],sizeof(b[1]),0); 62 for i:=n-1 downto 1 do 63 begin 64 while (a[k,1]-m)>a[i,1] do dec(k); 65 if getmax(i+1,k)>=(a[i,2]*2) then b[i,2]:=1; 66 end; 67 l:=0;for i:=1 to n do inc(l,(b[i,1]+b[i,2]) div 2); 68 writeln(l); 69 readln; 70 71 end.