codeforce 599C Day at the Beach

Day at the Beach

 

limit 2s 256M

 

    One day Squidward, Spongebob and Patrick decided to go to the beach. Unfortunately, the weather was bad, so the friends were unable to ride waves. However, they decided to spent their time building sand castles.

    At the end of the day there were n castles built by friends. Castles are numbered from 1 to n, and the height of the i-th castle is equal to hi. When friends were about to leave, Squidward noticed, that castles are not ordered by their height, and this looks ugly. Now friends are going to reorder the castles in a way to obtain that condition hi ≤ hi+1 holds for all i from 1 to n - 1.

Squidward suggested the following process of sorting castles:

    Castles are split into blocks — groups of consecutive castles. Therefore the block from i to j will include castles i,i+1,...,j. A block may consist of a single castle.

    The partitioning is chosen in such a way that every castle is a part of exactly one block.

Each block is sorted independently from other blocks, that is the sequence hi,hi + 1, ..., hj becomes sorted.

    The partitioning should satisfy the condition that after each block is sorted, the sequence hi becomes sorted too. This may always be achieved by saying that the whole sequence is a single block.

    Even Patrick understands that increasing the number of blocks in partitioning will ease the sorting process. Now friends ask you to count the maximum possible number of blocks in a partitioning that satisfies all the above requirements.

 

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 100 000) — the number of castles Spongebob, Patrick and Squidward made from sand during the day.

The next line contains n integers hi (1 ≤ hi ≤ 109). Thei-th of these integers corresponds to the height of the i-th castle

 

Output

Print the maximum possible number of blocks in a valid partitioning.

 

Example

Input

4

2 1 3 2

Output

2

 

題意

給定一序列，將其分組，使各組組內排序完成之後整個序列也是有序的

 

題解

    對於第i個組，顯然滿足max(i)<=min(i+1)

    但是這個條件沒有什麼卵用……

    再想一想，假設目前已經在劃分i組，如果後面的元素裏有aj<max(i)，顯然必須把aj併入第i組，也就是對於第k個元素，可不可以在這裏劃分取決於k之後的元素中的最小值是否大於等於k之前的那一組數的最大值。

再往深一步，就是比較k之前的最大值和k之後的最小值。

數列前綴後綴的最值都可以O(n)求出

#include<stdio.h>
#include<iostream>
#include<algorithm>

using namespace std;
typedef struct my_pair my_pair;
const int maxn=100000;

struct my_pair
{
	int extreme;
	int x;
};

int max_x[maxn+5];
my_pair min_x[maxn+5];

int arr[maxn+5];

int main(void)
{
	int n;
	scanf("%d",&n);
	
	for (int i=1;i<=n;++i)
	{
		scanf("%d",&arr[i]);
	}
	
	min_x[n].extreme=arr[n];
	min_x[n].x=n;
	for (int i=n-1;i>=1;--i)
	{
		if (arr[i]>min_x[i+1].extreme)
		{
			min_x[i]=min_x[i+1];
		}
		else
		{
			min_x[i].extreme=arr[i];
			min_x[i].x=i;
		}
	}
	
	max_x[1]=arr[1];
	for (int i=2;i<=n;++i)
	{
		max_x[i]=max(max_x[i-1],arr[i]);
	}
	
	int ans=2;
	for (int i=2;i<=n;++i)
	{
		if (max_x[i-1]<=min_x[i].extreme) ++ans;
		else
		{
			i=min_x[i].x;
		}
	}
	
	printf("%d\n",ans-1);
}