P2947 [USACO09MAR]仰望Look Up
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- 74通过
- 122提交
- 题目提供者洛谷OnlineJudge
- 标签USACO2009云端
- 难度普及/提高-
- 时空限制1s / 128MB
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- 中文翻译应当为向右看齐
- 题目中文版范围。。
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.
分析：和前几题差不多，不过就是要记录一下每次栈顶的位置.
#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
int n,h[100010],ans[100010],stk[100010],top,num[100010];
int main()
{
scanf("%d", &n);
for (int i = 1; i <= n; i++)
scanf("%d", &h[i]);
for (int i = n; i >= 1; i--)
{
while (top != 0 && stk[top] <= h[i])
top--;
ans[i] = num[top];
stk[++top] = h[i];
num[top] = i;
}
for (int i = 1; i <= n; i++)
printf("%d\n", ans[i]);
return 0;
}