POJ3250[USACO2006Nov]Bad Hair Day[单调栈]

Bad Hair Day

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 17774		Accepted: 6000

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

6
10
3
7
4
12
2

Sample Output

5

Source

USACO 2006 November Silver

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题意：求每个点右面能看到的个数的总和

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超级水的单调栈

注意两个相同高度就看不到了

//
//  main.cpp
//  poj3250
//
//  Created by Candy on 10/6/16.
//  Copyright © 2016 Candy. All rights reserved.
//

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long ll;
const int N=8e4+5;
inline int read(){
    char c=getchar();int x=0,f=1;
    while(c<'0'||c>'9'){if(c=='-')f=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*f;
}
int n,a[N];
ll ans=0;
struct data{
    int h,c;
}st[N];
int top=0;
int main(int argc, const char * argv[]) {
    n=read();
    for(int i=n;i>=1;i--) a[i]=read();
    for(int i=1;i<=n;i++){
        data x;
        x.h=a[i]; x.c=0;
        while(top&&st[top].h<x.h){
            x.c+=st[top].c+1;
            top--;
        }
        st[++top]=x;
        ans+=x.c;
        //printf("%d %d\n",i,x.c);
    }
    printf("%lld",ans);
    return 0;
}