bzoj2338

计算几何

我们先把所有的线段求出来，我们发现只有两个线段等长且中点重合时才能构成矩形，那么线段有n*n条，我们按中点，长度排序，然后对于一条线段扫描所有符合条件的线段计算答案，这样看起来是O(n^3)次的，实际上远远到不了

但是1336和1765两道题空间较小，不能乱开空间

#include<bits/stdc++.h>
using namespace std;
const int N = ;
int n;
double ans;
int x[N], y[N];
struct line {
    int mx, my, x1, x2, y1, y2;
    long long len;
    bool friend operator < (line A, line B) {
        if(A.mx != B.mx) return A.mx < B.mx;
        if(A.my != B.my) return A.my < B.my;
        if(A.len != B.len) return A.len < B.len;
        return true;
    }
    line(int mx, int my, long long len, int x1, int y1, int x2, int y2) : mx(mx), my(my), len(len), x1(x1), y1(y1), x2(x2), y2(y2) {}
};
vector<line> v;
inline long long sqr(long long x) { return x * x; }
inline double Area(line a, line b)
{
    double disa = sqrt(sqr(a.x1 - b.x1) + sqr(a.y1 - b.y1)), disb = sqrt(sqr(a.x2 - b.x1) + sqr(a.y1 - b.y2));
    return disa * disb;
}
int main()
{
//  freopen("crectangle.in", "r", stdin);
//  freopen("crectangle.out", "w", stdout);
    scanf("%d", &n);
    for(int i = ; i <= n; ++i) scanf("%d%d", &x[i], &y[i]);
    for(int i = ; i <= n; ++i)
        for(int j = i + ; j <= n; ++j)
        {
            long long len = sqr(x[i] - x[j]) + sqr(y[i] - y[j]);
            v.push_back(line(x[i] + x[j], y[i] + y[j], len, x[i], y[i], x[j], y[j]));
        }
    sort(v.begin(), v.end());
    for(int i = ; i < v.size(); ++i)
    {
        line a = v[i];
        for(int j = i - ; j >= ; --j)
        {
            line b = v[j];
            if(a.mx != b.mx || a.my != b.my || a.len != b.len) break;
            ans = max(ans, Area(a, b));
        }
    }
    printf("%.0f\n", ans);
//  fclose(stdin);
//  fclose(stdout);
    return ;
}