POJ 3250：Bad Hair Day 好玩的单调栈

Bad Hair Day

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 15699		Accepted: 5255

Description

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 h_i (1 ≤ h_i ≤ 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 c_i denote the number of cows whose hairstyle is visible from cow i; please compute the sum of c₁ through c_N.For this example, the desired is answer 3 + 0 + 1 + 0 + 1 + 0 = 5.

Input

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.

Output

Line 1: A single integer that is the sum of c₁ through c_N.

Sample Input

6
10
3
7
4
12
2

Sample Output

5

题意是站着一排牛，牛从左往右看能看到比自己身高小的牛的发型，但是如果碰到牛的身高比自己高了，那么到此为止。

转换一下思维，想象一下牛是从右往左看只能看到比自己身高大的牛，然后如果身高变小，那么到此为止。

很好玩的题目，用单调栈来做，从左往右读，栈内元素从栈底到栈顶是递增的，这样如果遇到元素比栈顶元素大，那么弹出。然后计算此时栈内元素的个数，相加即得到结果。

代码：

#include <iostream>
#include <algorithm>
#include <cmath>
#include <vector>
#include <string>
#include <cstring>
#pragma warning(disable:4996)
using namespace std;

#define N 80002

long long a[N], stack[N], top;

int main()
{
	//freopen("i.txt", "r", stdin);
	//freopen("o.txt", "w", stdout);

	long long ans, tmp;
	int i, j, n;

	scanf("%d", &n);

	for (i = 1; i <= n; i++)
	{
		scanf("%lld", a + i);
	}
	a[++n] = 1e9 + 7;
	top = 0;
	ans = 0;
	for (i = 1; i <= n; i++)
	{
		while (top >= 1 && a[i] >= a[stack[top - 1]])
		{
			--top;
		}
		ans = ans + top;
		if (top == 0 || a[i] < a[stack[top - 1]])
		{
			stack[top++] = i;
		}
	}
	printf("%lld\n", ans);

	//system("pause");
	return 0;
}

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