BZOJ1185 [HNOI2007]最小矩形覆盖

  1 /**************************************************************
  2     Problem: 1185
  3     User: rausen
  4     Language: C++
  5     Result: Accepted
  6     Time:180 ms
  7     Memory:2372 kb
  8 ****************************************************************/
  9  
 10 #include <cstdio>
 11 #include <cmath>
 12 #include <algorithm>
 13  
 14 #define P Point
 15 using namespace std;
 16 typedef double lf;
 17 const int N = 50005;
 18 const lf eps = 1e-8;
 19  
 20 struct Point {
 21     lf x, y;
 22     P() {}
 23     P(lf _x, lf _y) : x(_x), y(_y) {}
 24      
 25     inline bool operator < (const P &X) const {
 26         return fabs(y - X.y) < eps ? x < X.x : y < X.y;
 27     }
 28     inline bool operator == (const P &X) const {
 29         return fabs(x - X.x) < eps && fabs(y - X.y) < eps;
 30     }
 31     inline bool operator != (const P &X) const {
 32         return !(*this == X);
 33     }
 34     inline P operator + (const P &X) const {
 35         return P(x + X.x, y + X.y);
 36     }
 37     inline P operator - (const P &X) const {
 38         return P(x - X.x, y - X.y);
 39     }
 40     inline P operator * (const lf &X) const {
 41         return P(x * X, y * X);
 42     }
 43     inline lf operator * (const P &X) const {
 44         return x * X.y - y * X.x;
 45     }
 46     inline lf operator / (const P &X) const {
 47         return x * X.x + y * X.y;
 48     }
 49     inline void read() {
 50         scanf("%lf%lf", &x, &y);
 51     }
 52 }p[N], s[N], t[5];
 53  
 54 int n, top;
 55 lf ans = 1e60;
 56  
 57 inline lf sqr(lf x) {
 58     return x * x;
 59 }
 60  
 61 inline lf dis(P a, P b) {
 62     return sqrt(sqr(a.x - b.x) + sqr(a.y - b.y));
 63 }
 64  
 65 inline bool cmp_p(P a, P b) {
 66     lf tmp = (a - p[1]) * (b - p[1]);
 67     return fabs(tmp) < eps ? dis(a, p[1]) - dis(b, p[1]) < eps : tmp > -eps;
 68 }
 69  
 70 void graham() {
 71     int i;
 72     for (i = 2; i <= n; ++i)
 73         if (p[i] < p[1]) swap(p[1], p[i]);
 74     sort(p + 2, p + n + 1, cmp_p);
 75     for (i = 2, s[top = 1] = p[1]; i <= n; ++i) {
 76         while (top > 1 && (s[top] - s[top - 1]) * (p[i] - s[top]) < eps) --top;
 77         s[++top] = p[i];
 78     }
 79     s[0] = s[top];
 80 }
 81  
 82 inline bool check_p(int i, int p) {
 83     return (s[i + 1] - s[i]) * (s[p + 1] - s[i]) - (s[i + 1] - s[i]) * (s[p] - s[i]) > -eps;
 84 }
 85  
 86 inline bool check_r(int i, int r) {
 87     return (s[i + 1] - s[i]) / (s[r + 1] - s[i]) - (s[i + 1] - s[i]) / (s[r] - s[i]) > -eps;
 88 }
 89  
 90 inline bool check_l(int i, int l) {
 91     return (s[i + 1] - s[i]) / (s[l + 1] - s[i]) - (s[i + 1] - s[i]) / (s[l] - s[i]) < eps;
 92 }
 93  
 94 void work() {
 95     int l, r, p, i;
 96     lf L, R, H, D, tmp;
 97     for (i = 0, l = r = p = 1; i < top; ++i) {
 98         D = dis(s[i], s[i + 1]);
 99         while (check_p(i, p)) (p += 1) %= top;
100         while (check_r(i, r)) (r += 1) %= top;
101         if (!i) l = r;
102         while (check_l(i, l)) (l += 1) %= top;
103         L = (s[i + 1] - s[i]) / (s[l] - s[i]) / D;
104         R = (s[i + 1] - s[i]) / (s[r] - s[i]) / D;
105         H = (s[i + 1] - s[i]) * (s[p] - s[i]) / D;
106         if (H < 0) H = -H;
107         tmp = (R - L) * H;
108         if (tmp < ans) {
109             ans = tmp;
110             t[0] = s[i] + (s[i + 1] - s[i]) * (R / D);
111             t[1] = t[0] + (s[r] - t[0]) * (H / dis(t[0], s[r]));
112             t[2] = t[1] - (t[0] - s[i]) * ((R - L) / dis(s[i], t[0]));
113             t[3] = t[2] - (t[1] - t[0]);
114         }
115     }
116 }
117  
118 int main() {
119     int i, st;
120     scanf("%d", &n);
121     for (i = 1; i <= n; ++i)
122         p[i].read();
123     graham();
124     work();
125     for (i = 1, st = 0; i <= 3; ++i)
126         if (t[i] < t[st]) st = i;
127     for (i = 0, printf("%.5lf\n", ans); i <= 3; ++i)
128         printf("%.5lf %.5lf\n", t[(i + st) % 4].x, t[(i + st) % 4].y);
129     return 0;
130 }