• bzoj 1185


    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
    ", ans);
    116     for(int i = 0; i < 4; i++) {
    117         printf("%.5f %.5f
    ", vec2[i].x, vec2[i].y);
    118     }
    119     return 0;
    120 }
    121 /*
    122 */
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  • 原文地址:https://www.cnblogs.com/CJLHY/p/9020078.html
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