# AtCoder - 4351 Median of Medians(二分+线段树求顺序对)

D - Median of Medians

Time limit : 2sec / Memory limit : 1024MB

Score : 700 points
Problem Statement

We will define the median of a sequence b of length M, as follows:

Let b' be the sequence obtained by sorting b in non-decreasing order. Then, the value of the (M⁄2+1)-th element of b' is the median of b. Here, ⁄ is integer division, rounding down.

For example, the median of (10,30,20) is 20; the median of (10,30,20,40) is 30; the median of (10,10,10,20,30) is 10.

Snuke comes up with the following problem.

You are given a sequence a of length N. For each pair (l,r) (1≤l≤r≤N), let ml,r be the median of the contiguous subsequence (al,al+1,…,ar) of a. We will list ml,r for all pairs (l,r) to create a new sequence m. Find the median of m.
Constraints

1≤N≤105
ai is an integer.
1≤ai≤109

Input

Input is given from Standard Input in the following format:

N
a1 a2 … aN

Output

Print the median of m.
Sample Input 1
Copy

3
10 30 20

Sample Output 1
Copy

30

The median of each contiguous subsequence of a is as follows:

The median of (10) is 10.
The median of (30) is 30.
The median of (20) is 20.
The median of (10,30) is 30.
The median of (30,20) is 30.
The median of (10,30,20) is 20.

Thus, m=(10,30,20,30,30,20) and the median of m is 30.
Sample Input 2
Copy

1
10

Sample Output 2
Copy

10

Sample Input 3
Copy

10
5 9 5 9 8 9 3 5 4 3

Sample Output 3
Copy

8

#include<bits/stdc++.h>
#define lson root<<1
#define rson root<<1|1
using namespace std;

int n;
int a[200010];
int sum[200010];

struct node
{
int l,r,sum;
}tr[1600080];

int push_up(int root)
{
tr[root].sum=tr[lson].sum+tr[rson].sum;
}

int build(int root,int l,int r)
{
if(l==r)
{
tr[root].sum=0;
tr[root].l=l;
tr[root].r=r;
return 0;
}
tr[root].l=l;
tr[root].r=r;
int mid=(l+r)>>1;
build(lson,l,mid);
build(rson,mid+1,r);
push_up(root);
}

int update(int root,int pos)
{
if(pos==tr[root].l&&pos==tr[root].r)
{
tr[root].sum++;
return 0;
}
int mid=(tr[root].l+tr[root].r)>>1;
if(pos<=mid)
{
update(lson,pos);
}
else
{
update(rson,pos);
}
push_up(root);
}

int query(int root,int l,int r)
{
if(l>r) return 0;
if(l<=tr[root].l&&tr[root].r<=r) return tr[root].sum;
int mid=(tr[root].l+tr[root].r)>>1;
if(r<=mid)
{
return query(lson,l,r);
}
else
{
if(l>mid)
{
return query(rson,l,r);
}
else
{
return query(lson,l,mid)+query(rson,mid+1,r);
}
}
}

int check(int x)
{
long long ans=0;
memset(sum,0,sizeof(sum));
for(int i=1;i<=n;i++)
{
sum[i]=sum[i-1]+((a[i]<=x)?1:-1);
}
for(int i=0;i<=n;i++)
{
sum[i]+=n+1;
}
build(1,1,200010);
for(int i=0;i<=n;i++)
{
ans+=query(1,1,sum[i]-1);
update(1,sum[i]);
}
return ans>=1ll*n*(n+1)/4+1;
}

int main()
{
scanf("%d",&n);
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
}
int l=1,r=1e9,mid;
while(l<=r)
{
mid=(l+r)>>1;
if(check(mid))
{
r=mid;
}
else
{
l=mid+1;
}
if(r-l<=1)
{
mid=check(l)?l:r;
break;
}
}
printf("%d\n",mid);
}

posted @ 2018-10-07 16:43  Styx-ferryman  阅读(632)  评论(0编辑  收藏  举报