Codeforces Round #332 (Div. 2) C. Day at the Beach 线段树

C. Day at the Beach

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/599/problem/C

Description

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 tohi. 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). The i-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.

Sample Input

3
1 2 3

Sample Output

3

HINT

题意

给你n个数，问你最多分成多少块，使得在块内单独排序之后，然后再组合起来的结果和最后排序的结果一样

题解：

对于每个数i，判断一下这个数到上一个划开的位置的最大值是否小于后面的最小值，如果是，就在这儿划开

最大最小值可以On，也可以偷懒像我一样用线段树。。。

代码

#include<iostream>
#include<stdio.h>
using namespace std;

#define maxn 500000
const int inf = 1e9+;
struct node
{
    int l,r,ma,mi;
}a[maxn];
int d[maxn];
void build(int x,int l,int r)
{
    a[x].l=l,a[x].r=r;
    a[x].ma=-inf,a[x].mi=inf;
    if(l==r)
    {
        a[x].ma=d[l];
        a[x].mi=d[l];
        return;
    }
    else
    {
        int mid=(l+r)>>;
        build(x<<,l,mid);
        build(x<<|,mid+,r);
        a[x].ma=max(a[x<<].ma,a[x<<|].ma);
        a[x].mi=min(a[x<<].mi,a[x<<|].mi);
    }
}
int query1(int x,int st,int ed)
{
    int l=a[x].l,r=a[x].r;
    if(st<=l&&r<=ed)
        return a[x].mi;
    else
    {
        int mid=(l+r)>>;
        int mi1=inf,mi2=inf;
        if(st<=mid)
            mi1=query1(x<<,st,ed);
        if(ed>mid)
            mi2=query1(x<<|,st,ed);
        return min(mi1,mi2);
    }
}

int query2(int x,int st,int ed)
{
    int l=a[x].l,r=a[x].r;
    if(st<=l&&r<=ed)
        return a[x].ma;
    else
    {
        int mid=(l+r)>>;
        int mi1=-inf,mi2=-inf;
        if(st<=mid)
            mi1=query2(x<<,st,ed);
        if(ed>mid)
            mi2=query2(x<<|,st,ed);
        return max(mi1,mi2);
    }
}

int main()
{
    int n;
    scanf("%d",&n);
    int dd,ee;
    for(int i=;i<=n;i++)
        scanf("%d",&d[i]);
    build(,,n);
    int ans = ;
    int tmp = ;
    for(int i=;i<n;i++)
    {
        if(query2(,tmp,i)<=query1(,i+,n))
        {
            ans++;
            tmp = i+;
        }
    }
    cout<<ans<<endl;
}