POJ 3250 Bad Hair Day (单调栈应用)_some of farmer john's n cows (1 ≤ n ≤ 80,000) are -优快云博客

农夫约翰的N头奶牛中有些正在经历糟糕的发型日。每头奶牛的高度为hi，面对东方站立。FJ想要计算一头奶牛能看到其他奶牛头顶的数量。如果一头奶牛前面的奶牛比它矮，那么它就能看到那头奶牛的头顶。题目要求计算每头奶牛能看到的其他奶牛头顶数量之和。例如，给定6头奶牛的高度，计算结果为5。输入包含奶牛的数量和每头奶牛的高度，输出为所有奶牛能看到的奶牛头顶数之和。

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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

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 c 1 through cN.

Sample Input

Sample Output

#include<iostream>
#include<algorithm>
#include<string>
#include<map>//int dx[4]={0,0,-1,1};int dy[4]={-1,1,0,0};
#include<queue>//int gcd(int a,int b){return b?gcd(b,a%b):a;}
#include<vector>
#include<cmath>
#include<stack>
#include<string.h>
#include<stdlib.h>
#include<cstdio>
#define mod 1e9+7
#define ll unsigned long long
#define MAX 1000000000
#define ms memset
const int maxn=80005;
using namespace std;

int n;
int cnt[maxn],l,r;///单调栈的结构
/*
题目大意：给定一个序列代表一排母牛的高度，
要求每个牛朝右能看到的牛的数量的和（只能看到比自己矮的）。

单调栈维护着递增序列，
对每个进来的数，其删除的数全部记录，
但不仅是要记录个数，实际上还要计数
该数对应的动态规划的值。
用结构体记录下标即可。

*/

struct node
{
    int id,h;
};
node s[maxn],sk[maxn];

int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&s[i].h);
        s[i].id=i;
    }

    l=0,r=-1;
    for(int i=n;i>=1;i--)
    {
        int res=0,tp=0;
        while(r>=l&&s[i].h>sk[r].h)
        {
            tp++;
            res+=cnt[sk[r].id];
            r--;
        }
        cnt[i]=res+tp;
        sk[++r]=s[i];
    }

    ll ans=0;
    for(int i=1;i<=n;i++ )   ans+=cnt[i];
    printf("%lld\n",ans);

    return 0;
}