poj 3250 Bad Hair Day(单调栈)
http://poj.org/problem?id=3250
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 ≤ 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
Lines 2..N+1: Line i+1 contains a single integer that is the height of cow i.
Output
Sample Input
6 10 3 7 4 12 2
Sample Output
5
一群高度不完全相同的牛从左到右站成一排，每头牛只能看见它右边的比它矮的牛的发型，若遇到一头高度大于或等于它的牛，则无法继续看到这头牛后面的其他牛。
给出这些牛的高度，要求每头牛可以看到的牛的数量的和。
把要求作一下转换，其实就是要求每头牛被看到的次数之和。这个可以使用单调栈来解决。
从左到右依次读取当前牛的高度，从栈顶开始把高度小于或等于当前牛的高度的那些元素删除，此时栈中剩下的元素的数量就是可以看见当前牛的其他牛的数量。把这个数量加在一起，就可以得到最后的答案了。
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <limits>
#include <queue>
#include <stack>
#include <vector>
#include <map>
using namespace std;
typedef long long LL;
#define N 100000
#define INF 0x3f3f3f3f
#define PI acos (-1.0)
#define EPS 1e-8
#define met(a, b) memset (a, b, sizeof (a))
int main ()
{
int n, num;
while (scanf ("%d", &n) != EOF)
{
int top = 0, Stack[N];
LL ans = 0;
while (n--)
{
scanf ("%d", &num);
while (top > 0 && Stack[top-1] <= num)
top--;
ans += top;
Stack[top++] = num;
}
printf ("%lld\n", ans);
}
return 0;
}