poj3250Bad Hair Day【单调栈】

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Bad Hair Day

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 16176 Accepted: 5481

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

610374122

Sample Output

5

题意：一群牛有高有低站成一队每个牛可以看到它右边比他个低的牛如果有一个牛比它高则此后的牛都看不到；求每个牛能看到的牛的数量和

思路：把每个牛能看到牛的个数转化为能看到当前牛的个数。用单调栈；

#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#include<cmath>#include<queue>#include<list>#include<vector>using namespace std;const int maxn=80010;long long stack[maxn]; int main(){int n,i,j,k;while(scanf("%d",&n)!=EOF){int top=0;long long ans=0,num; for(i=1;i<=n;++i){scanf("%d",&num);while(top>0&&stack[top-1]<=num)top--;ans+=top;stack[top++]=num;}printf("%lld\n",ans);}return 0;}