uva 12307(点集的外接矩形)

题意：平面上有n个点，要求求出包含所有点的矩形的最小面积和最小周长。
题解：先求点集的凸包，然后把凸包的每条边当做底边，把其他左、上、右三边根据底边用旋转卡壳的方式确定，计算长宽更新最小值。做了这题后对旋转卡壳理解更深刻了。

#include <cstdio>#include <cstring>#include <cmath>#include <vector>#include <algorithm>using namespace std;const double eps = 1e-11;const double PI = acos(-1);int dcmp(double x) {    if (fabs(x) < eps)        return 0;    return x > 0 ? 1 : -1;}struct Point {    double x, y;    Point (double a = 0, double b = 0): x(a), y(b) {}};typedef Point Vector;typedef vector<Point> Polygon;Vector operator + (const Vector& a, const Vector& b) { return Vector(a.x + b.x, a.y + b.y); }Vector operator - (const Vector& a, const Vector& b) { return Vector(a.x - b.x, a.y - b.y); }Vector operator * (const Vector& a, double& b) { return Vector(a.x * b, a.y * b); }Vector operator / (const Vector& a, double& b) { return Vector(a.x / b, a.y / b); }bool operator == (const Vector& a, const Vector& b) { return !dcmp(a.x - b.x) && !dcmp(a.y - b.y); }bool operator < (const Vector& a, const Vector& b) { return a.x < b.x || (a.x == b.x && a.y < b.y); }double Dot(const Vector& a, const Vector& b) { return a.x * b.x + a.y * b.y; }double Length(const Vector& a) { return sqrt(Dot(a, a)); }double Length2(const Vector& a) { return Dot(a, a); }double Cross(const Vector& a, const Vector& b) { return a.x * b.y - a.y * b.x; }double Angle(const Vector& a, const Vector& b) { return acos(Dot(a, b) / Length(a) / Length(b)); }Vector Rotate(Vector A, double rad) { return Point(A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad)); }int ConvexHull(Point* P, int cnt, Point* res) {    sort(P, P + cnt);    cnt = unique(P, P + cnt) - P;    int m = 0;    for (int i = 0; i < cnt; i++) {        while (m > 1 && Cross(res[m - 1] - res[m - 2], P[i] - res[m - 2]) <= 0)            m--;        res[m++] = P[i];    }    int k = m;    for (int i = cnt - 2; i >= 0; i--) {        while (m > k && Cross(res[m - 1] - res[m - 2], P[i] - res[m - 2]) <= 0)            m--;        res[m++] = P[i];    }    if (cnt > 1)        m--;    return m;}void GetMinRCarea(Point* P, int cnt, double& S, double& C) {    S = C = 1e18;    int l = 1, r = 1, u = 1;    for (int i = 0; i < cnt; i++) { //枚举底边P[i]~P[i+1]        while (Cross(P[i + 1] - P[i], P[u + 1] - P[u]) > eps)            u = (u + 1) % cnt;//找上边，夹角还能增加就向后找        while (Dot(P[i + 1] - P[i], P[r + 1] - P[r]) > eps)             r = (r + 1) % cnt;//找右边，夹角还小于90°就继续找        if (i == 0)            l = (r + 1) % cnt;        while (Dot(P[i + 1] - P[i], P[l + 1] - P[l]) < -eps)            l = (l + 1) % cnt;//找左边，夹角还大于90°就继续找        double d = Length(P[i + 1] - P[i]); //底边长度        double a = Cross(P[i + 1] - P[i], P[u] - P[i]) / d;//叉积几何意义就是求向量组成的平行四边形面积，除以底边就是高        double b = (Dot(P[r] - P[i], P[i + 1] - P[i]) - Dot(P[l] - P[i], P[i + 1] - P[i])) / d;//向左向右的投影和(点积几何意义：A*B = |A||B|cosθ一个向量在另一个向量方向上的投影长度乘后者的长度)，所以最后还要除后者的长度。        S = min(S, a * b);        C = min(C, 2.0 * (a + b));    }}const int N = 100005;Point P[N], R[N];int n;int main() {    while(scanf("%d", &n) == 1 && n) {        for (int i = 0; i < n; i++)            scanf("%lf%lf", &R[i].x, &R[i].y);        int cnt = ConvexHull(R, n, P);        P[cnt] = P[0];        double S, C;        GetMinRCarea(P, cnt, S, C);        printf("%.2lf %.2lf\n", S, C);    }    return 0;}