Time Limit: 3 Sec Memory Limit: 128 MB
Submit: 136 Solved: 81
约翰的N(1≤N≤105)头奶牛站成一排,奶牛i的身高是Hi(l≤Hi≤1,000,000).现在,每只奶牛都在向左看齐.对于奶牛i,如果奶牛j满足i<j且Hi<Hj,我们可以说奶牛i可以仰望奶牛j. 求出每只奶牛离她最近的仰望对象.
第1行输入N,之后每行输入一个身高.
共N行,按顺序每行输出一只奶牛的最近仰望对象.如果没有仰望对象,输出0.
6 3 2 6 1 1 2
3 3 0 6 6 0
题解:再一次请出我神奇的小线段树——用线段树实现求在某某位置之后大于某值的最靠左位置,用了一个比较神奇的分治,具体如程序(发现线段是是个乱搞神器啊有木有)
1 var
2 i,j,k,l,m,n:longint;
3 a,b:array[0..1000000] of longint;
4 function max(x,y:longint):longint;inline;
5 begin
6 if x>y then max:=x else max:=y;
7 end;
8 function min(x,y:longint):longint;inline;
9 begin
10 if x<y then min:=x else min:=y;
11 end;
12 procedure built(z,x,y:longint);inline;
13 begin
14 if x=y then
15 begin
16 read(a[z]);
17 b[x]:=a[z];
18 end
19 else
20 begin
21 built(z*2,x,(x+y) div 2);
22 built(z*2+1,(x+y) div 2+1,y);
23 a[z]:=max(a[z*2],a[z*2+1]);
24 end;
25 end;
26 function approach(z,x,y,l,r,t:longint):longint;inline;
27 var a1:longint;
28 begin
29 if l>r then exit(0);
30 if a[z]<=t then exit(0);
31 if x=y then
32 begin
33 if a[z]>t then exit(x);
34 end;
35 a1:=approach(z*2,x,(x+y) div 2,l,min(r,(x+y) div 2),t);
36 if a1<>0 then exit(a1);
37 exit(approach(z*2+1,(x+y) div 2+1,y,max((x+y) div 2+1,l),r,t));
38 end;
39 begin
40 readln(n);
41 built(1,1,n);
42 for i:=1 to n do
43 writeln(approach(1,1,n,i+1,n,b[i]));
44
45 end.