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.

题解：

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn=100000+5;typedef long long ll;int read(){int x=0,f=1;char ch;ch=getchar();while(ch<'0'||ch>'9'){if(ch=='-') f=-1; ch=getchar();}while(ch>='0'&&ch<='9'){x=x*10+ch-'0'; ch=getchar();}return x*f;}int n;int s[maxn],h[maxn],a[maxn];int main(){int i,top=1;n=read();h[1]=read();s[1]=1;for(i=2;i<=n;i++){h[i]=read();while(top&&h[i]>h[s[top]]) {a[s[top]]=i;top--;}s[++top]=i;}for(i=1;i<=n;i++)printf("%d\n",a[i]);return 0;}