Acwing-121- cattle into the ring (half, and two-dimensional prefix, discretized)

link:

https://www.acwing.com/problem/content/123/

Meaning of the questions:

Farmer John wanted to build a corral for his cows.

These critical requirements corral animals must be square and must contain at least clover C units, to put out as their afternoon tea.

Corral edges must be parallel to the X, Y axis.

John's land contains a total of N unit clover, clover per unit land area is located within a 1 x 1, the location area represented by the lower left corner coordinates, and a corner of the zone X, Y coordinates are integers in the range 1 to 10000 or less.

Clover plurality of units may be located within the same region of a 1 x 1, and for this reason, the next input, may occur once in the same area coordinates.

Only one of the region located entirely repaired corral, only I think clover in this region in the corral.

The case where you help John calculate, can contain at least C unit area clover, minimum side length of the corral is.

Ideas:

After a two-dimensional discrete and do some prefixes and again to the query can be reduced to about 500 ^ 2 * log (1000).

Code:

#include <bits/stdc++.h>
using namespace std;

struct Node
{
    int x, y;
    int node;
}node[510];
int Map[10000][10000];
int Number[2010];
int c, n, pos;

bool Check(int len)
{
    for (int x1 = 1, x2 = 1;x2 <= pos;x2++)
    {
        while (Number[x2]-Number[x1]+1 > len)
            x1++;
        for (int y1 = 1, y2 = 1;y2 <= pos;y2++)
        {
            while (Number[y2]-Number[y1]+1 > len)
                y1++;
            if (Map[x2][y2]-Map[x2][y1-1]-Map[x1-1][y2]+Map[x1-1][y1-1] >= c)
                return true;
        }
    }
    return false;
}

int main()
{
    scanf("%d %d", &c, &n);
    int x, y;
    pos = 0;
    for (int i = 1;i <= n;i++)
    {
        scanf("%d%d", &node[i].x, &node[i].y);
        Number[++pos] = node[i].x;
        Number[++pos] = node[i].y;
    }
    sort(Number+1, Number+1+2*n);
    pos = unique(Number+1, Number+1+2*n)-(Number+1);
    for (int i = 1;i <= n;i++)
    {
        node[i].x = lower_bound(Number+1, Number+1+pos, node[i].x)-Number;
        node[i].y = lower_bound(Number+1, Number+1+pos, node[i].y)-Number;
        Map[node[i].x][node[i].y]++;
    }
    for (int i = 1;i <= pos;i++)
    {
        for (int j = 1;j <= pos;j++)
            Map[i][j] = Map[i][j]+Map[i-1][j]+Map[i][j-1]-Map[i-1][j-1];
    }
    int l = 1, r = 10000;
    int res = 10000;
    while (l < r)
    {
//        cout << l << ' ' << r << endl;
        int mid = (l+r)/2;
        if (Check(mid))
        {
            r = mid;
        }
        else
            l = mid+1;
    }
//    cout << r << endl;
    printf("%d\n", r);

    return 0;
}
/*
9 9
1 1
1 2
1 3
2 1
2 2
2 3
3 1
3 2
1000 1000
 */