3314: [Usaco2013 Nov]Crowded Cows

3314: [Usaco2013 Nov]Crowded Cows

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 111  Solved: 79
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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

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：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

/**************************************************************
    Problem: 3314
    User: HansBug
    Language: Pascal
    Result: Accepted
    Time:608 ms
    Memory:6008 kb
****************************************************************/
 
var
   i,j,k,l,m,n:longint;
   a,b:array[0..100000,1..2] of longint;
   c:array[0..17,0..60000] of longint;
function max(x,y:longint):longint;
         begin
              if x>y then max:=x else max:=y;
         end;
procedure swap(var x,y:longint);
          var z:longint;
          begin
               z:=x;x:=y;y:=z;
          end;
procedure sort(l,r:longint);
          var i,j,x:longint;
          begin
               i:=l;j:=r;x:=a[(l+r) div 2,1];
               repeat
                     while a[i,1]<x do inc(i);
                     while a[j,1]>x do dec(j);
                     if i<=j then
                        begin
                             swap(a[i,1],a[j,1]);
                             swap(a[i,2],a[j,2]);
                             inc(i);dec(j);
                        end;
               until i>j;
               if i<r then sort(i,r);
               if l<j then sort(l,j);
          end;
function getmax(x,y:longint):longint;
         var i:longint;
         begin
              if y<x then exit(-1);
              i:=trunc(ln(y-x+1)/ln(2));
              exit(max(c[i,x],c[i,y-trunc(exp(ln(2)*i))+1]));
         end;
begin
     readln(n,m);
     for i:=1 to n do readln(a[i,1],a[i,2]);
     sort(1,n);
     for i:=1 to n do c[0,i]:=a[i,2];
     for i:=1 to trunc(ln(n)/ln(2)+1) do
         for j:=1 to n-trunc(exp(ln(2)*i))+1 do
             c[i,j]:=max(c[i-1,j],c[i-1,j+trunc(exp(ln(2)*(i-1)))]);
     k:=1;fillchar(b[1],sizeof(b[1]),0);
     for i:=2 to n do
         begin
              while (a[k,1]+m)<a[i,1] do inc(k);
              if getmax(k,i-1)>=(a[i,2]*2) then b[i,1]:=1;
         end;
     k:=n;fillchar(b[1],sizeof(b[1]),0);
     for i:=n-1 downto 1 do
         begin
              while (a[k,1]-m)>a[i,1] do dec(k);
              if getmax(i+1,k)>=(a[i,2]*2) then b[i,2]:=1;
         end;
     l:=0;for i:=1 to n do inc(l,(b[i,1]+b[i,2]) div 2);
     writeln(l);
     readln;
 
end.