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:
     User: HansBug
     Language: Pascal
     Result: Accepted
     Time: ms
     Memory: kb
 ****************************************************************/

 var
    i,j,k,l,m,n:longint;
    a,b:array[..,..] of longint;
    c:array[..,..] 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 ,];
                repeat
                      while a[i,]<x do inc(i);
                      while a[j,]>x do dec(j);
                      if i<=j then
                         begin
                              swap(a[i,],a[j,]);
                              swap(a[i,],a[j,]);
                              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(-);
               i:=trunc(ln(y-x+)/ln());
               exit(max(c[i,x],c[i,y-trunc(exp(ln()*i))+]));
          end;
 begin
      readln(n,m);
      for i:= to n do readln(a[i,],a[i,]);
      sort(,n);
      for i:= to n do c[,i]:=a[i,];
      for i:= to trunc(ln(n)/ln()+) do
          for j:= to n-trunc(exp(ln()*i))+ do
              c[i,j]:=max(c[i-,j],c[i-,j+trunc(exp(ln()*(i-)))]);
      k:=;fillchar(b[],sizeof(b[]),);
      for i:= to n do
          begin
               while (a[k,]+m)<a[i,] do inc(k);
               if getmax(k,i-)>=(a[i,]*) then b[i,]:=;
          end;
      k:=n;fillchar(b[],sizeof(b[]),);
      for i:=n- downto  do
          begin
               while (a[k,]-m)>a[i,] do dec(k);
               if getmax(i+,k)>=(a[i,]*) then b[i,]:=;
          end;
      l:=;for i:= to n do inc(l,(b[i,]+b[i,]) div );
      writeln(l);
      readln;

 end.