BZOJ2338 [HNOI2011]数矩形

  1 /**************************************************************
  2     Problem: 2338
  3     User: rausen
  4     Language: C++
  5     Result: Accepted
  6     Time:1808 ms
  7     Memory:36224 kb
  8 ****************************************************************/
  9  
 10 #include <cstdio>
 11 #include <algorithm>
 12  
 13 #define P Point
 14 #define L Line
 15 using namespace std;
 16 typedef long long ll;
 17 const int N = 1505;
 18 const int M = N * N >> 1;
 19  
 20 int n, tot;
 21 ll ans;
 22  
 23 struct Point {
 24     ll x, y;
 25     P() {}
 26     P(ll _x, ll _y) : x(_x), y(_y) {}
 27      
 28     inline bool operator == (const P &b) const {
 29         return x == b.x && y == b.y;
 30     }
 31     inline bool operator < (const P &b) const {
 32         return x == b.x ? y < b.y : x < b.x;
 33     }
 34     inline P operator + (const P &b) const {
 35         return P(x + b.x, y + b.y);
 36     }
 37     inline P operator - (const P &b) const {
 38         return P(x - b.x, y - b.y);
 39     }
 40     inline ll operator * (const P &b) const {
 41         return x * b.y - y * b.x;
 42     }
 43 } a[N];
 44  
 45 struct Line {
 46     int a, b;
 47     ll len;
 48     P mid;
 49     L() {}
 50     L(int _a, int _b, ll _l, P _m) : a(_a), b(_b), len(_l), mid(_m) {}
 51      
 52     inline bool operator == (const L &b) const {
 53         return len == b.len && mid == b.mid;
 54     }
 55     inline bool operator < (const L &b) const {
 56         return len == b.len ? mid < b.mid : len < b.len;
 57     }
 58 } l[M];
 59  
 60 inline ll read() {
 61     ll x = 0, sgn = 1;
 62     char ch = getchar();
 63     while (ch < ‘0‘ || ‘9‘ < ch) {
 64         if (ch == ‘-‘) sgn = -1;
 65         ch = getchar();
 66     }
 67     while (‘0‘ <= ch && ch <= ‘9‘) {
 68         x = x * 10 + ch - ‘0‘;
 69         ch = getchar();
 70     }
 71     return sgn * x;
 72 }
 73  
 74 inline ll sqr(ll x) {
 75     return (ll) x * x;
 76 }
 77  
 78 inline ll dist(P a, P b) {
 79     return (ll) sqr(a.x - b.x) + sqr(a.y - b.y);
 80 }
 81  
 82 inline ll abs_ll(ll x) {
 83     return x < 0 ? -x : x;
 84 }
 85  
 86 int main() {
 87     int i, j;
 88     n = read();
 89     for (i = 1; i <= n; ++i)
 90         a[i].x = read(), a[i].y = read();
 91     for (i = 1; i < n; ++i)
 92         for (j = i + 1; j <= n; ++j)
 93             l[++tot] = L(i, j, dist(a[i], a[j]), a[i] + a[j]);
 94     sort(l + 1, l + tot + 1);
 95     for (i = 1; i <= tot; ++i)
 96         for (j = i - 1; j && l[i] == l[j]; --j)
 97             ans = max(ans, abs_ll((a[l[i].a] - a[l[j].a]) * (a[l[i].a] - a[l[j].b])));
 98     printf("%lld\n", ans);
 99     return 0;
100 }