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 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
Lines 2..N+1: Line i+1 contains a single integer that is the height of cow i.
Output
Sample Input
6
10
3
7
4
12
2
Sample Output
5
Source
题意:求每个点右面能看到的个数的总和
超级水的单调栈
//
// 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+;
inline int read(){
char c=getchar();int x=,f=;
while(c<''||c>''){if(c=='-')f=-;c=getchar();}
while(c>=''&&c<=''){x=x*+c-'';c=getchar();}
return x*f;
}
int n,a[N];
ll ans=;
struct data{
int h,c;
}st[N];
int top=;
int main(int argc, const char * argv[]) {
n=read();
for(int i=n;i>=;i--) a[i]=read();
for(int i=;i<=n;i++){
data x;
x.h=a[i]; x.c=;
while(top&&st[top].h<x.h){
x.c+=st[top].c+;
top--;
}
st[++top]=x;
ans+=x.c;
//printf("%d %d\n",i,x.c);
}
printf("%lld",ans);
return ;
}