P2866 [USACO06NOV]糟糕的一天Bad Hair Day--单调栈

P2866 [USACO06NOV]糟糕的一天Bad Hair Day

题意翻译

农夫约翰有N (N leq 80000)N(N≤80000)头奶牛正在过乱头发节。每一头牛都站在同一排面朝东方，而且每一头牛的身高为h_ihi​。第NN头牛在最前面，而第11头牛在最后面。 对于第ii头牛前面的第jj头牛，如果h_i>h_{i+1}hi​>hi+1​并且h_i>h_{i+2}hi​>hi+2​ cdots⋯ h_i>h_jhi​>hj​，那么认为第ii头牛可以看到第i+1i+1到第jj头牛

定义C_iCi​为第ii头牛所能看到的别的牛的头发的数量。请帮助农夫约翰求出sum_{i=1}^n C_i∑i=1n​Ci​

题目描述

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.

输入输出格式

输入格式：

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.

输出格式：

Line 1: A single integer that is the sum of c1 through cN.

输入输出样例

输入样例#1： 复制

6
10
3
7
4
12
2

输出样例#1： 复制

5
单调栈经典题，首先我们维护一个严格递减的单调栈，当我们读入一个新元素时，如果这个新元素小于栈顶元素，就入栈，否则就弹出栈顶元素，并且ans加上两个元素下标之差-1（可以画图看看），同时我们还应该在最后赋一个极大值来将栈中所有的元素弹出，这样问题就解决了，记得开long long。

#include<iostream>
#include<cstdio>
#include<string>
#include<cmath>
#include<cstring>
#include<queue>
#include<stack>
#include<algorithm>
#define maxn 80005
using namespace std;
stack<int>s;

inline int read()
{
    char c=getchar();
    int res=0,x=1;
    while(c<'0'||c>'9')
    {
        if(c=='-')
        x=-1;
        c=getchar();
    }
    while(c>='0'&&c<='9')
    {
        res=res*10+(c-'0');
        c=getchar();
    }
    return x*res;
}

long long ans;
int n,aa;
long long a[maxn];

int main()
{
    n=read();
    for(int i=1;i<=n;i++)
    {
        aa=read();
        a[i]=aa;
    }
    a[n+1]=10000000005;//赋成极大值 
    for(int i=1;i<=n+1;i++)
    {
        if(s.empty()||a[i]<a[s.top()])//维护一个单调递减栈 
        {
            s.push(i);
        }
        else 
        {
            while(!s.empty()&&a[i]>=a[s.top()])
            {
                ans+=(long long)(i-s.top()-1);//加上两个元素的下标之差-1 
                s.pop();
            }
            s.push(i);
        }
    }
    printf("%lld",ans);
    return 0;
}

To the world you may be one person, but to one person you may be the world.
对于世界而言，你是一个人；但是对于某个人，你是他的整个世界。

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