poj 3250 Bad Hair Day

Posted 2020-11-03 ZefengYao

Bad Hair Day

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 21084		Accepted: 7202

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：个子高的牛能看到个子低的牛，且每头牛的视线都是一个方向,问每头牛分别能看到视线方向上的多少头牛。
思路：单调栈,从后往前考虑队列,若当前的牛能看到存储在栈中的牛，则把栈中的牛从栈中抛出,直到栈为空或者栈中的牛的身高大于等于当前牛的身高为止，将当前的牛压入栈中，每次执行这样的操作即可。
AC代码：

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<algorithm>
#include<vector>
#include<cstring>
#include<string>
#include<cmath>
using namespace std;
#define INF 0x3f3f3f3f
#define N_MAX 80000+20
typedef long long ll;
int n,h[N_MAX];
int st[N_MAX],num[N_MAX];
int main() {
    scanf("%d",&n);
    for (int i = 0; i < n; i++)scanf("%d",&h[i]);
    int t = 0;//栈的大小
    for (int i = n-1; i >=0; i--) {
        while (t > 0 && h[st[t - 1]] <h[i])t--;
        num[i] = t == 0 ? n - 1 - i : st[t - 1] - i - 1;
        st[t++]= i;
    }
    ll sum = 0;
    for (int i = 0; i < n; i++)
        sum += num[i];
    printf("%lld\n",sum);
    return 0;
}