【POJ 3250】牛的视野

Bad Hair Day

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 25050		Accepted: 8525

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

Source

USACO 2006 November Silver

题解：计算每头牛可以看到牛的头数，可以等效于每头牛可以被看到的次数

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，求和就是答案

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <stack>
#include <string>
#include <cmath>
#include <cstdio>
#include <functional> 
typedef long long ll;
#define N 1005
using namespace std;
 
int main()
{
    int n;
    while(scanf("%d",&n)==1){
        int num;
        ll sum=0;
        scanf("%d",&num);
        stack<int>s;
        while(!s.empty())
            s.pop();
        s.push(num);
        for(int i=1;i<n;i++){
            scanf("%d",&num);
            while(!s.empty()&&num>=s.top()) //当前牛小于栈顶
                s.pop();
            sum+=s.size();
            s.push(num);
        }
        printf("%lld\n",sum);
    }
    return 0;