bzoj 1185

Posted 2020-11-03 notnight

题目大意： 给你n个点求最小矩形覆盖。

 

思路：枚举凸包上的边然后，旋转卡壳找三个相应的为止把矩形的四个点求出来。

  1 #include<bits/stdc++.h>
  2 #define LL long long
  3 #define fi first
  4 #define se second
  5 #define mk make_pair
  6 #define pii pair<int,int>
  7 #define piii pair<int, pair<int,int>>
  8 
  9 using namespace std;
 10 
 11 const int N=1e5 + 7;
 12 const int M=1e4 + 7;
 13 const int inf = 0x3f3f3f3f;
 14 const LL INF = 0x3f3f3f3f3f3f3f3f;
 15 const int mod = 1e9 + 7;
 16 const double eps = 1e-10;
 17 const double PI = acos(-1);
 18 
 19 int n, cnt;
 20 
 21 int dcmp(double x) {
 22     if(fabs(x) < eps) return 0;
 23     else return x < 0 ? -1 : 1;
 24 }
 25 
 26 struct Point {
 27     double x, y;
 28     Point(double x = 0, double y = 0) : x(x), y(y) { }
 29 
 30 }p[N], ch[N];
 31 
 32 typedef Point Vector;
 33 
 34 Point operator + (Vector A, Vector B) {return Point(A.x + B.x, A.y + B.y);}
 35 Point operator - (Vector A, Vector B) {return Point(A.x - B.x, A.y - B.y);}
 36 Point operator * (Vector A, double p) {return Point(A.x * p, A.y * p);}
 37 Point operator / (Vector A, double p) {return Point(A.x / p, A.y / p);}
 38 bool operator < (const Vector &A, const Vector &B) {return A.y < B.y || (A.y == B.y && A.x < B.x);}
 39 bool operator == (const Vector &A, const Point &B) {return dcmp(A.x - B.x) == 0 && dcmp(A.y - B.y) == 0;}
 40 double Dot(Vector A, Vector B) {return A.x * B.x + A.y * B.y;}
 41 double Length(Vector A) {return sqrt(Dot(A, A));}
 42 double Angle(Vector A, Vector B) {return acos(Dot(A, B) / Length(A) / Length(B));}
 43 double Cross(Vector A, Vector B) {return A.x * B.y - A.y * B.x;}
 44 double Area2(Point A, Point B, Point C) {return Cross(B - A, C - A);}
 45 
 46 Vector Rotate(Vector A, double rad) {
 47     return Vector(A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad));
 48 }
 49 
 50 Point GetLineIntersection(Point P, Vector v, Point Q, Vector w) {
 51     Vector u = P - Q;
 52     double t = Cross(w, u) / Cross(v, w);
 53     return P + v * t;
 54 }
 55 
 56 double dis(Point A, Point B) {
 57     return sqrt((A.x - B.x) * (A.x - B.x) + (A.y - B.y) * (A.y - B.y));
 58 }
 59 int ConvexHull(Point *p, int n, Point *ch) {
 60     sort(p, p + n);
 61     int m = 0;
 62     for(int i = 0; i < n; i++) {
 63         while(m > 1 && dcmp(Cross(ch[m - 1] - ch[m - 2], p[i] - ch[m - 2])) <= 0) m--;
 64         ch[m++] = p[i];
 65     }
 66 
 67     int k = m;
 68     for(int i = n - 2; i >= 0; i--) {
 69         while(m > k && dcmp(Cross(ch[m - 1] - ch[m - 2], p[i] - ch[m - 2])) <= 0) m--;
 70         ch[m++] = p[i];
 71     }
 72     return m;
 73 }
 74 
 75 Point vec[10], vec2[10];
 76 
 77 int main() {
 78     scanf("%d", &n);
 79     for(int i = 0; i < n; i++)
 80         scanf("%lf%lf", &p[i].x, &p[i].y);
 81     cnt = ConvexHull(p, n, ch);
 82 
 83     cnt--;
 84     for(int i = 0; i < cnt; i++) {
 85         ch[cnt + i] = ch[i];
 86     }
 87 
 88     int pos1 = 1, pos2 = 1, pos3 = 1;
 89     double ans = inf;
 90     for(int i = 0; i < cnt; i++) {
 91         while(abs(Cross(ch[i] - ch[pos1 + 1], ch[i + 1] - ch[pos1 + 1])) > abs(Cross(ch[i] - ch[pos1], ch[i + 1] - ch[pos1])))
 92             pos1++;
 93         while(Dot(ch[i + 1] - ch[i], ch[pos2 + 1] - ch[i]) > Dot(ch[i + 1] - ch[i], ch[pos2] - ch[i]))
 94             pos2++;
 95         pos3 = max(pos3, pos1);
 96         while(Dot(ch[i + 1] - ch[i], ch[pos3 + 1] - ch[i]) < Dot(ch[i + 1] - ch[i], ch[pos3] - ch[i]))
 97             pos3++;
 98         Vector k1 = ch[i + 1] - ch[i];
 99         Vector k2 = Rotate(k1, PI / 2);
100         Point p1 = GetLineIntersection(ch[i], k1, ch[pos2], k2);
101         Point p2 = GetLineIntersection(ch[i], k1, ch[pos3], k2);
102         Point p3 = GetLineIntersection(ch[pos1], k1, ch[pos2], k2);
103         Point p4 = GetLineIntersection(ch[pos1], k1, ch[pos3], k2);
104         double ret = dis(p1, p2) * dis(p1, p3);
105         if(ret < ans) {
106             ans = ret;
107             vec[0] = p1;
108             vec[1] = p2;
109             vec[2] = p3;
110             vec[3] = p4;
111         }
112     }
113 
114     ConvexHull(vec, 4, vec2);
115     printf("%.5f\n", ans);
116     for(int i = 0; i < 4; i++) {
117         printf("%.5f %.5f\n", vec2[i].x, vec2[i].y);
118     }
119     return 0;
120 }
121 /*
122 */