【POJ3250】Bad Hair Day

本文介绍了一道名为BadHairDay的编程题,该题要求计算每头牛能看到前方多少头比它矮的牛的数量,并使用单调栈算法进行解答。通过样例解释了问题背景和求解思路。

Bad Hair Day
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 17576 Accepted: 5926

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 hi (1 ≤ h≤ 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.

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  c 1 through  cN.

Sample Input

6
10
3
7
4
12
2

Sample Output

5

Source

用单调栈求解。

从栈底到栈顶单调递减栈,可以看成每头牛被看到多少次。

#include<cstdio>
#include<iostream>
#include<stack>
#define LL long long
using namespace std;
int main() {
	int n;
	while(scanf("%d",&n)!=EOF) {
		stack<int >S;
		int L;
		LL ans=0;
		for(int i=0; i<n; i++) {
			scanf("%d",&L);
			while(!S.empty()&&S.top()<=L) {
				S.pop();
			}
			ans+=S.size();
			S.push(L);
		}
		printf("%lld\n",ans);
	}
	return 0;
}
题目地址:http://poj.org/problem?id=3250




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