问题 H: Pond(二维前缀和+二分）

The land of a park AtCoder is an N×N grid with east-west rows and north-south columns. The height of the square at the i-th row from the north and j-th column from the west is given as Ai,j.

Takahashi, the manager, has decided to build a square pond occupying K×K squares in this park.
To do this, he wants to choose a square section of K×K squares completely within the park whose median of the heights of the squares is the lowest. Find the median of the heights of the squares in such a section.

Here, the median of the heights of the squares in a K×K section is the height of the (⌊K2/2⌋+1)-th ighest square among the K2 squares in the section, where ⌊x⌋ is the greatest integer not exceeding x.

Constraints
1≤K≤N≤800
0≤Ai,j≤109
All values in input are integers.

输入

Input is given from Standard Input in the following format:
N K
A1,1 A1,2 … A1,N
A2,1 A2,2 … A2,N
:
AN,1 AN,2 … AN,N

输出

Print the answer.

样例输入 Copy

【样例1】
3 2
1 7 0
5 8 11
10 4 2
【样例2】
3 3
1 2 3
4 5 6
7 8 9

样例输出 Copy

【样例1】
4
【样例2】
5

提示

样例1解释：
Let (i,j) denote the square at the i-th row from the north and j-th column from the west. We have four candidates for the 2×2 section occupied by the pond: 
{(1,1),(1,2),(2,1),(2,2)},{(1,2),(1,3),(2,2),(2,3)},{(2,1),(2,2),(3,1),(3,2)},{(2,2),(2,3),(3,2),(3,3)}.
When K=2, since ⌊22/2⌋+1=3, the median of the heights of the squares in a section is the height of the 3-rd highest square, which is 5, 7, 5, 4 for the candidates above, respectively. We should print the lowest of these: 4.

 

 

 

 

 

题意

给出一个  阶方阵，计算所有  阶方阵的中位数的最小值。

这里  阶方阵的中位数为：  个数中第  个大的数。

题解

二分中位数的值，将大于二分值的数记为  ，小于等于的记为  ，作二维前缀和来枚举所有  阶方阵。

如果存在一个方阵满足其和小于等于  ，则当前二分值可行，设为上界，否则设为下界。

 

二分+二维前缀和

#include<iostream>
#include<algorithm>
#include<vector>
using namespace std;
const int maxn=1e3+100;
int a[maxn][maxn];
int n,k;
int judge(int x){
    vector<vector<int>>sum(n+1,vector<int>(n+1));
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            sum[i][j]=sum[i-1][j]+sum[i][j-1]-sum[i-1][j-1]+(a[i][j]>x);
        } 
    }
    for(int i=k;i<=n;i++){
        for(int j=k;j<=n;j++){
            int z=sum[i][j]-sum[i-k][j]-sum[i][j-k]+sum[i-k][j-k];
            if(z<=k*k/2){
                return 1;
            }
        }
    }
    return 0;
}
int main(){
    cin>>n>>k;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            scanf("%d",&a[i][j]);
        }
    }
    int l=0,r=1e9,ans=0;
    while(r>=l){
        int mid=(l+r)/2;
        if(judge(mid)){
            ans=mid;
            r=mid-1;
        }
        else{
            l=mid+1;
        }
    }
    cout<<ans<<endl;
}