IndiaHacks 2016 - Online Edition (Div. 1 + Div. 2) C. Bear and Up-Down 暴力

C. Bear and Up-Down

题目连接：

http://www.codeforces.com/contest/653/problem/C

Description

The life goes up and down, just like nice sequences. Sequence t1, t2, ..., tn is called nice if the following two conditions are satisfied:

ti < ti + 1 for each odd i < n;
ti > ti + 1 for each even i < n.
For example, sequences (2, 8), (1, 5, 1) and (2, 5, 1, 100, 99, 120) are nice, while (1, 1), (1, 2, 3) and (2, 5, 3, 2) are not.

Bear Limak has a sequence of positive integers t1, t2, ..., tn. This sequence is not nice now and Limak wants to fix it by a single swap. He is going to choose two indices i < j and swap elements ti and tj in order to get a nice sequence. Count the number of ways to do so. Two ways are considered different if indices of elements chosen for a swap are different.

Input

The first line of the input contains one integer n (2 ≤ n ≤ 150 000) — the length of the sequence.

The second line contains n integers t1, t2, ..., tn (1 ≤ ti ≤ 150 000) — the initial sequence. It's guaranteed that the given sequence is not nice.

Output

Print the number of ways to swap two elements exactly once in order to get a nice sequence.

Sample Input

5
2 8 4 7 7

Sample Output

2

Hint

题意

一个序列定义为nice的话，就是这个序列满足阶梯状。

就是如果i是偶数，那么ai>ai-1,ai>ai+1

如果i是奇数，那么ai<ai+1,ai<ai-1

现在允许你交换两个数的位置，问你一共有多少种交换方式，使得这个序列变成nice

保证一开始的序列不是nice的。

题解：

我们定义不nice的数就是不满足条件的位置。

我们可以大胆猜测一发，不nice的数一定不会有很多，因为一次交换最多影响6个数，所以我们把这些不nice的数直接扔到一个数组里面。

然后暴力去和整个序列去交换就好了。

然后check也是只用check那些不nice的位置和你交换的那个位置的数。

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 2e5+7;
int n;
int a[maxn];
vector<int>tmp;
long long ans = 0;
set<pair<int,int> >S;
bool check()
{
    for(int i=0;i<tmp.size();i++)
    {
        for(int j=-1;j<=1;j++)
        {
            if(tmp[i]+j==0)continue;
            if(tmp[i]+j==n+1)continue;
            int pos = (tmp[i]+j);
            if(pos%2==1)
            {
                if(a[pos]>=a[pos+1])return false;
                if(a[pos]>=a[pos-1])return false;
            }
            else
            {
                if(a[pos]<=a[pos+1])return false;
                if(a[pos]<=a[pos-1])return false;
            }
        }
    }
    return true;
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    a[0]=1e9;
    if(n%2==1)a[n+1]=1e9;
    else a[n+1]=-1;
    for(int i=1;i<=n;i++)
    {
    	if( i & 1 ){
			bool ok = true;
			if( i + 1 <= n && a[i] >= a[i+1] ) ok = false;
			if( i - 1 >= 1 && a[i] >= a[i-1] ) ok = false;
			if( ok == false ) tmp.push_back( i );
		}else{
			bool ok = true;
			if( i + 1 <= n && a[i] <= a[i+1] ) ok = false;
			if( i - 1 >= 1 && a[i] <= a[i-1] ) ok = false;
			if( ok == false ) tmp.push_back( i );
		}
    }

    if(tmp.size()>30)
        return puts("0"),0;

    for(int i=0;i<tmp.size();i++)
    {
        for(int j=1;j<=n;j++)
        {
            if(tmp[i]==j)continue;
            swap(a[tmp[i]],a[j]);
            bool ok = check();
            if(j%2==1)
            {
                if(a[j]>=a[j+1]||a[j]>=a[j-1])ok=false;
            }
            if(j%2==0)
            {
                if(a[j]<=a[j+1]||a[j]<=a[j-1])ok=false;
            }
            if(ok)
            {
                pair < int , int > SS = make_pair( min( tmp[i] , j ) , max( tmp[i] , j ) );
                if(!S.count(SS)){
                    S.insert( SS );
                    ans++;
                }
            }
            swap(a[tmp[i]],a[j]);
        }
    }
    cout<<ans<<endl;
}