洛谷 2947 仰望Look Up(单调栈)

P2947 [USACO09MAR]仰望Look Up

Farmer John’s N (1 <= N <= 100,000) cows, conveniently numbered 1..N, are once again standing in a row. Cow i has height H_i (1 <= H_i <= 1,000,000).

Each cow is looking to her left toward those with higher index numbers. We say that cow i ‘looks up’ to cow j if i < j and H_i < H_j. For each cow i, FJ would like to know the index of the first cow in line looked up to by cow i.

Note: about 50% of the test data will have N <= 1,000.

约翰的N(1≤N≤10^5)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小于j且Hi小于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．

Input

输入输出格式

输入格式：
Line 1: A single integer: N

Lines 2..N+1: Line i+1 contains the single integer: H_i
输出格式：
Lines 1..N: Line i contains a single integer representing the smallest index of a cow up to which cow i looks. If no such cow exists, print 0.
输入输出样例

输入样例#1：
6
3
2
6
1
1
2
输出样例#1：
3
3
0
6
6
0
说明

FJ has six cows of heights 3, 2, 6, 1, 1, and 2.

Cows 1 and 2 both look up to cow 3; cows 4 and 5 both look up to cow 6; and cows 3 and 6 do not look up to any cow.

program df;
var i,j,n,m,x,y,z,k,t:longint;
a,b,c:array[0..1000000] of longint;
begin
readln(n);
for i:=1 to n do
readln(c[i]);
t:=0; a[0]:=0;
for i:=1 to n do
begin
while (t>0) and (c[i]>c[b[t]]) do //维护单调栈
begin
a[b[t]]:=i;
dec(t);
end;
inc(t);
b[t]:=i;
end;
for i:=1 to n do
writeln(a[i]);
end.