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=1nCi
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
6 10 3 7 4 12 2
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。
1 #include<iostream> 2 #include<cstdio> 3 #include<string> 4 #include<cmath> 5 #include<cstring> 6 #include<queue> 7 #include<stack> 8 #include<algorithm> 9 #define maxn 80005 10 using namespace std; 11 stack<int>s; 12 13 inline int read() 14 { 15 char c=getchar(); 16 int res=0,x=1; 17 while(c<'0'||c>'9') 18 { 19 if(c=='-') 20 x=-1; 21 c=getchar(); 22 } 23 while(c>='0'&&c<='9') 24 { 25 res=res*10+(c-'0'); 26 c=getchar(); 27 } 28 return x*res; 29 } 30 31 long long ans; 32 int n,aa; 33 long long a[maxn]; 34 35 int main() 36 { 37 n=read(); 38 for(int i=1;i<=n;i++) 39 { 40 aa=read(); 41 a[i]=aa; 42 } 43 a[n+1]=10000000005;//赋成极大值 44 for(int i=1;i<=n+1;i++) 45 { 46 if(s.empty()||a[i]<a[s.top()])//维护一个单调递减栈 47 { 48 s.push(i); 49 } 50 else 51 { 52 while(!s.empty()&&a[i]>=a[s.top()]) 53 { 54 ans+=(long long)(i-s.top()-1);//加上两个元素的下标之差-1 55 s.pop(); 56 } 57 s.push(i); 58 } 59 } 60 printf("%lld",ans); 61 return 0; 62 }
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。
--snowy 2019-01-18 14:06:50