P2947 [USACO09MAR]向右看齐Look Up（单调栈）

题目描述
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，每只奶牛都在向右看齐．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．

Input

输入格式

• Line 1: A single integer: N

• Lines 2…N+1: Line i+1 contains the single integer: H_i

第 1 行输入 N，之后每行输入一个身高 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.

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。

输入输出样例
输入 #1复制
6
3
2
6
1
1
2
输出 #1复制
3
3
0
6
6
0
说明/提示
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.

【输入说明】6 头奶牛的身高分别为 3, 2, 6, 1, 1, 2.

【输出说明】奶牛#1,#2 仰望奶牛#3，奶牛#4,#5 仰望奶牛#6，奶牛#3 和#6 没有仰望对象。

【数据规模】

对于 20%的数据： 1≤N≤10；

对于 50%的数据： 1≤N≤1,000；

对于 100%的数据：1≤N≤100,000；1≤H_i≤1,000,000；

水。。。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long ll;

const int maxn = 1e6 + 7;

int a[maxn],num[maxn];
struct STK
{
    int i,h;
}stk[maxn];

int main()
{
    int n;scanf("%d",&n);
    for(int i = 1;i <= n;i++)
    {
        scanf("%d",&a[i]);
    }
    
    int top = 0;
    for(int i = 1;i <= n;i++)//递减单调栈
    {
        while(top && stk[top].h < a[i])
        {
            num[stk[top].i] = i;
            top--;
        }
        stk[++top].i = i;stk[top].h = a[i];
    }
    for(int i = 1;i <= n;i++)printf("%d\n",num[i]);
    return 0;
}