P2866 [USACO06NOV]糟糕的一天Bad Hair Day
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- 75通过
- 153提交
- 题目提供者洛谷OnlineJudge
- 标签USACO2006云端
- 难度普及/提高-
- 时空限制1s / 128MB
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- 题目标题
题目描述
Some of Farmer John‘s N cows (1 ≤ N ≤ 80,000) are having a bad hair day! Since each cow is self-conscious about her messy hairstyle, FJ wants to count the number of other cows that can see the top of other cows‘ heads.
Each cow i has a specified height hi (1 ≤ hi ≤ 1,000,000,000) and is standing in a line of cows all facing east (to the right in our diagrams). Therefore, cow i can see the tops of the heads of cows in front of her (namely cows i+1, i+2, and so on), for as long as these cows are strictly shorter than cow i.
Consider this example:
=
= =
= - = Cows facing right -->
= = =
= - = = =
= = = = = =
1 2 3 4 5 6 Cow#1 can see the hairstyle of cows #2, 3, 4
Cow#2 can see no cow‘s hairstyle
Cow#3 can see the hairstyle of cow #4
Cow#4 can see no cow‘s hairstyle
Cow#5 can see the hairstyle of cow 6
Cow#6 can see no cows at all!
Let ci denote the number of cows whose hairstyle is visible from cow i; please compute the sum of c1 through cN.For this example, the desired is answer 3 + 0 + 1 + 0 + 1 + 0 = 5.
农民约翰的某N(1 < N < 80000)头奶牛正在过乱头发节！由于每头牛都意识到自己凌乱不堪 的发型，约翰希望统计出能够看到其他牛的头发的牛的数量.
每一头牛i有一个高度所有N头牛面向东方排成一排，牛N在最前面，而 牛1在最后面.第i头牛可以看到她前面的那些牛的头，只要那些牛的高度严格小于她的高度，而且 中间没有比hi高或相等的奶牛阻隔.
让N表示第i头牛可以看到发型的牛的数量；请输出Ci的总和
输入输出格式
输入格式：
Line 1: The number of cows, N.
Lines 2..N+1: Line i+1 contains a single integer that is the height of cow i.
输出格式：
Line 1: A single integer that is the sum of c1 through cN.
输入输出样例
输入样例#1：
6 10 3 7 4 12 2
输出样例#1：
5分析：很显然是单调栈，至于单调栈的用法可以去看我的前几篇博客，这里计数器累加的top代表这只牛的头发能被前面的多少只牛看到（因为前面有top只牛比这只牛高）.
#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
int n,h[500010],stk[500010],top;
long long ans;
int main()
{
scanf("%d", &n);
for (int i = 1; i <= n; i++)
scanf("%d", &h[i]);
for (int i = 1; i <= n; i++)
{
while (top != 0 && stk[top] <= h[i])
top--;
ans += top;
stk[++top] = h[i];
}
printf("%lld", ans);
return 0;
}