洛谷 P2947 [USACO09MAR]仰望Look Up

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．

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.

栈。

屠龙宝刀点击就送

#include <cstdio>
#define N 100005
int a[N],num[N],ans[N],n,last,stack[N],top;
int main()
{
    scanf("%d",&n);
    for(int i=;i<=n;i++)
    {
        scanf("%d",&a[i]);num[i]=i;
        if(!top) {stack[++top]=i;continue;}
        else if(a[stack[top]]<a[i])
        {
            for(;(a[stack[top]]<a[i])&&top;top--)
            ans[num[stack[top]]]=i;
        }
        stack[++top]=i;
    }
    for(int i=;i<=n;i++) printf("%d\n",ans[i]);
    return ;
}