NEFU 635(二分+枚举)

题目：Twinkle Twinkle Little Star

题意：就是给n个点的坐标，然后在这个图形中找出一个边长最小的正方形，要求正方形的边平行于坐标轴且覆盖的点大于等于k个。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

const int N = 200010;
const int INF = 999999999;

struct Point
{
    int x,y;
    bool operator < (const Point &a) const
    {
         if(x!=a.x) return x < a.x;
         return y < a.y;
    }
};

Point p[N];

int n,k,y[N];

int judge(int len)
{
    int pre = -INF;
    for(int i=0,j;i<n;i++)
    {
        if(pre!=p[i].x)
        {
             pre = p[i].x;
             int cnt = 0;
             for(j=i;j<n;j++)
             {
                 if(p[j].x>=pre&&p[j].x<=pre+len)
                     y[cnt++] = p[j].y;
                 else break;
             }
             sort(y,y+cnt);
             if(cnt<k) continue;
             int pre1 = -INF;
             for(int r=0;r+k-1<cnt;r++)
             {
                  if(pre1!=y[r])
                  {
                      pre1 = y[r];
                      int up = upper_bound(y,y+cnt,pre1+len)-y-r;
                      if(up>=k) return 1;
                  }
             }
        }
    }
    return 0;
}

int main()
{
    int t=1;
    while(~scanf("%d%d",&n,&k))
    {
        for(int i=0;i<n;i++)
            scanf("%d%d",&p[i].x,&p[i].y);
        sort(p,p+n);

        int R = max(p[n-1].x-p[0].x,p[n-1].y-p[0].y);
        int L = 0;
        int ret = 0;
        while(L<=R)
        {
            int mid = (L+R)>>1;
            if (judge(mid))
            {
                ret = mid;
                R = mid-1;
            }
            else L = mid+1;
        }
        printf("Case %d: %d\n",t++,ret);
    }
    return 0;
}