Codeforces Beta Round #5 E. Bindian Signalizing 并查集

E. Bindian Signalizing

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/problemset/problem/5/E

Description

Everyone knows that long ago on the territory of present-day Berland there lived Bindian tribes. Their capital was surrounded by n hills, forming a circle. On each hill there was a watchman, who watched the neighbourhood day and night.

In case of any danger the watchman could make a fire on the hill. One watchman could see the signal of another watchman, if on the circle arc connecting the two hills there was no hill higher than any of the two. As for any two hills there are two different circle arcs connecting them, the signal was seen if the above mentioned condition was satisfied on at least one of the arcs. For example, for any two neighbouring watchmen it is true that the signal of one will be seen by the other.

An important characteristics of this watch system was the amount of pairs of watchmen able to see each other's signals. You are to find this amount by the given heights of the hills.

Input

The first line of the input data contains an integer number n (3 ≤ n ≤ 106), n — the amount of hills around the capital. The second line contains n numbers — heights of the hills in clockwise order. All height numbers are integer and lie between 1 and 109.

Output

Print the required amount of pairs.

Sample Input

5
1 2 4 5 3

Sample Output

7

HINT

 

题意

给你一个环的山，如果两个山之间没有比这座山更高的山的话，那么这两座山是可以互相看见的，然后问你最后有多少座山可以互相看见

题解：

首先我们把环变成一个链，然后用一个并查集的思想，处理每一座山左边第一个比这个山高的山是哪个，右边第一个比这个山高的是哪个，这个区域内和这个山高度一样的山有多少

然后跑一发并查集就好了

代码

 

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 1000010
#define mod 10007
#define eps 1e-9
int Num;
char CH[20];
//const int inf=0x7fffffff;
const int inf=0x3f3f3f3f;
/*

inline void P(int x)
{
    Num=0;if(!x){putchar('0');puts("");return;}
    while(x>0)CH[++Num]=x%10,x/=10;
    while(Num)putchar(CH[Num--]+48);
    puts("");
}
*/
inline ll read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
inline void P(int x)
{
    Num=0;if(!x){putchar('0');puts("");return;}
    while(x>0)CH[++Num]=x%10,x/=10;
    while(Num)putchar(CH[Num--]+48);
    puts("");
}
//**************************************************************************************

int a[maxn];
int b[maxn];
int l[maxn];
int r[maxn];
int k[maxn];
int main()
{
    int n=read();
    int p=-1;
    for(int i=0;i<n;i++)
    {
        a[i]=read();
        if(p==-1)
            p=i;
        else if(a[i]>a[p])
        {
            p=i;
        }
    }
    for(int i=0;i<n;i++)
        b[i]=a[(i+p+n)%n];
    b[n]=a[p];
    for(int i=1;i<n;i++)
    {
        l[i]=i-1;
        while(l[i]>0&&b[i]>=b[l[i]])
            l[i]=l[l[i]];
    }
    k[n]=0;
    for(int i=n-1;i>=0;i--)
    {
        r[i]=i+1;
        while(r[i]<n&&b[i]>b[r[i]])
            r[i]=r[r[i]];
        if(r[i]<n&&b[i]==b[r[i]])
        {
            k[i]=k[r[i]]+1;
            r[i]=r[r[i]];
        }
    }
    ll ans=0;
    for(int i=1;i<n;i++)
    {
        ans+=k[i];
        ans+=1;
        if(l[i]!=0||r[i]!=n)
            ans++;
    }
    cout<<ans<<endl;
}