[XSY 1947] Goodness

题意

　　给定 n 个点 $\left\{ p_i = (a_i, b_i) \right\}$ , 求 $\max_{i, j} |p_i \times p_j|$ .

　　n <= 100000 .

分析

　　凸包 + 三分 .

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <cstdlib>
 4 #include <cctype>
 5 #include <algorithm>
 6 using namespace std;
 7 #define F(i, a, b) for (register int i = (a); i <= (b); i++)
 8 #define LL long long
 9 inline LL rd(void) {
10     LL f = 1; char c = getchar(); for (; !isdigit(c); c = getchar()) if (c == ‘-‘) f = -1;
11     LL x = 0; for (; isdigit(c); c = getchar()) x = x*10+c-‘0‘; return x*f;
12 }
13 inline LL Abs(LL x) { return x < 0 ? -x : x; }
14
15 const int N = 200005;
16
17 int n, Len;
18 struct point {
19     LL x, y;
20     friend inline point operator - (point A, point B) { return (point){ A.x - B.x, A.y - B.y }; }
21     friend inline LL det(point A, point B) { return A.x * B.y - A.y * B.x; }
22 }Origin, p[N], s[N];
23 inline bool operator < (point A, point B) { return det(A - Origin, B - Origin) > 0; }
24
25 void Build(void) {
26     int ID = 1;
27     F(i, 2, n)
28         if (p[ID].y > p[i].y || (p[ID].y == p[i].y && p[ID].x > p[i].x))
29             ID = i;
30     swap(p[1], p[ID]);
31     sort(p+2, p+n+1);
32     F(i, 1, n) {
33         while (Len >= 2 && det(p[i] - s[Len-1], s[Len] - s[Len-1]) >= 0)
34             s[Len--] = (point){ 0, 0 };
35         s[++Len] = p[i];
36     }
37 }
38
39 inline LL Calc(point A, point B) { return Abs(det(A, B)); }
40 inline LL Search(point A, int l, int r) {
41     while (r-l > 3) {
42         int _l = (l + l + r) / 3; LL ansL = Calc(A, s[_l]);
43         int _r = (l + r + r) / 3; LL ansR = Calc(A, s[_r]);
44         ansL < ansR ? l = _l : r = _r;
45     }
46     LL res = 0;
47     F(i, l, r)
48         res = max(res, Calc(A, s[i]));
49     return res;
50 }
51
52 int main(void) {
53     #ifndef ONLINE_JUDGE
54         freopen("convex.in", "r", stdin);
55     #endif
56
57     n = rd();
58     F(i, 1, n) {
59         LL G = rd(), R = rd();
60         p[i] = (point){ G, R };
61     }
62
63     Build();
64     F(i, Len+1, Len+Len)
65         s[i] = s[i - Len];
66
67     LL res = 0;
68     F(i, 1, Len)
69         res = max(res, Search(s[i], i+1, i+Len));
70     printf("%lld\n", res);
71
72     return 0;
73 }