poj 3250 Bad Hair Day

最新推荐文章于 2020-02-05 23:24:01 发布

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最新推荐文章于 2020-02-05 23:24:01 发布

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原文链接：http://www.cnblogs.com/ZefengYao/p/9021025.html

本文详细解析了BadHairDay问题，这是一个经典的计算几何问题。通过使用单调栈算法，有效地解决了如何计算每头牛可以看到前方多少头比它矮的牛的问题，并给出了完整的AC代码。

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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

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;
}