POJ 3250 Bad Hair Day【单调栈】

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农夫约翰的N头奶牛中有些正经历糟糕的发型日。每头奶牛都有特定的高度hi，它们都面向东站立。FJ想要计算每头奶牛能看到多少其他奶牛的头顶。每头奶牛i能看到它前面所有比它矮的奶牛。使用单调栈解决此问题，求所有奶牛能看到的奶牛数量之和。

题目：

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.

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

题目大意：

n个牛排成一列向右看，牛 i 能看到牛 j 的头顶，当且仅当牛 j 在牛 i 的右边并且牛 i 与牛 j 之间的所有牛均比牛 i 矮。设牛 i 能看到的牛数为 Ci，求 ∑Ci ；

解题思路：

转换思路，从求每条牛能看到多少牛，转换成每条牛能被多少牛看到。

这样，只设置一个单调递减的栈即可，每有一头新牛加入栈时，入过它比栈顶牛高，那么栈顶的牛看不到这头牛，将栈顶的牛弹出，直到栈顶的牛比这头牛高，栈里剩余牛的数目，即为能看到这头牛的牛的数目。

实现代码：

#include <stack>
#include <cstdio>
#include <iostream>
#define MAXN 80007
using namespace std;

stack<int> st;
long long ans;
int n, h[MAXN], c[MAXN];

int main() {
	scanf("%d", &n);
	for(int i=1; i<=n; i++) {
		scanf("%d", &h[i]);
		while(!st.empty() && st.top()<=h[i]) {
			st.pop();
		}
		ans += st.size();
		st.push(h[i]);
	}
	printf("%lld", ans);
}