Bad Hair Day（问题抽象转化+“贡献值”思想+单调栈）

本文链接：https://blog.youkuaiyun.com/Enjoying_Science/article/details/47903141

Bad Hair Day

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 15528		Accepted: 5177

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 c ₁ through c_N.

Sample Input

Sample Output

Source

USACO 2006 November Silver

题意：n头牛从左到右排成一排，每头牛都向右看，且只能看到右边与其位置连续的所有高度比它小的牛，问所有牛能看到的牛的数目的总和是多少。

编程思想：题目问所有牛能看到的牛的数目的总和是多少，如果把问题对象仅仅局限在特定的某一头牛，然后按模拟的思想单独去求每头牛能看到右边与其位置连续的所有高度比它小的牛的数目，然后再加起来得到答案的话，效率不高，可能TLE。这时问题可以抽象转化一下，问所有牛能看到的牛的数目的总和是多少，其实等价于问每头牛对其左边的牛来说，左边有多少牛能看到它，那么它对该问题的贡献值就是多少。那么现在问题的关键是如何求每头牛对答案的贡献值呢？对于当前牛来说，左边的牛能看到它必须满足两个条件：左边的牛高度都比当前牛大，且左边这些高度比当前牛大的牛它们从左到右的高度只能是严格递减的。所以该思想可以利用“单调栈”这种数据结构来实现。在去除不满足条件的栈顶元素后，单调栈剩下的大小就是当前牛对最后答案的贡献值。这样直接扫描一遍，将所有贡献值加起来就是问题的答案了，时间复杂度为O(n)。

AC code：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <set>
#include <map>
#include <queue>
#define PI acos(-1.0)
#define LINF 1000000000000000000LL
#define eps 1e-8
#define LL long long
#define MAXN 1000010 
using namespace std;
int  st[MAXN];//单调栈 
LL ans;
int main()
{
	int i,j,n,ni,top;
	while(scanf("%d",&n)!=EOF)
	{
		ans=0;
		top=0;//栈顶指针（栈的大小） 
		for(i=1;i<=n;i++)
		{
			scanf("%d",&ni);
			while(top!=0&&ni>=st[top])
			{
				top--;//出栈 
			}
			ans+=top;//当前牛对最后答案的贡献值为栈的大小top 
			st[++top]=ni;//入栈 
		}
		printf("%lld\n",ans);
	}
    return 0;
}