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。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define pii pair<int,int>
#define piii pair<int, pair<int,int>>

using namespace std;

const int N=1e5 + 7;
const int M=1e4 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-10;
const double PI = acos(-1);

int n, cnt;

int dcmp(double x) {
    if(fabs(x) < eps) return 0;
    else return x < 0 ? -1 : 1;
}

struct Point {
    double x, y;
    Point(double x = 0, double y = 0) : x(x), y(y) { }

}p[N], ch[N];

typedef Point Vector;

Point operator + (Vector A, Vector B) {return Point(A.x + B.x, A.y + B.y);}
Point operator - (Vector A, Vector B) {return Point(A.x - B.x, A.y - B.y);}
Point operator * (Vector A, double p) {return Point(A.x * p, A.y * p);}
Point operator / (Vector A, double p) {return Point(A.x / p, A.y / p);}
bool operator < (const Vector &A, const Vector &B) {return A.y < B.y || (A.y == B.y && A.x < B.x);}
bool operator == (const Vector &A, const Point &B) {return dcmp(A.x - B.x) == 0 && dcmp(A.y - B.y) == 0;}
double Dot(Vector A, Vector B) {return A.x * B.x + A.y * B.y;}
double Length(Vector A) {return sqrt(Dot(A, A));}
double Angle(Vector A, Vector B) {return acos(Dot(A, B) / Length(A) / Length(B));}
double Cross(Vector A, Vector B) {return A.x * B.y - A.y * B.x;}
double Area2(Point A, Point B, Point C) {return Cross(B - A, C - A);}

Vector Rotate(Vector A, double rad) {
    return Vector(A.x * cos(rad) - A.y * sin(rad), A.x * sin(rad) + A.y * cos(rad));
}

Point GetLineIntersection(Point P, Vector v, Point Q, Vector w) {
    Vector u = P - Q;
    double t = Cross(w, u) / Cross(v, w);
    return P + v * t;
}

double dis(Point A, Point B) {
    return sqrt((A.x - B.x) * (A.x - B.x) + (A.y - B.y) * (A.y - B.y));
}
int ConvexHull(Point *p, int n, Point *ch) {
    sort(p, p + n);
    int m = 0;
    for(int i = 0; i < n; i++) {
        while(m > 1 && dcmp(Cross(ch[m - 1] - ch[m - 2], p[i] - ch[m - 2])) <= 0) m--;
        ch[m++] = p[i];
    }

    int k = m;
    for(int i = n - 2; i >= 0; i--) {
        while(m > k && dcmp(Cross(ch[m - 1] - ch[m - 2], p[i] - ch[m - 2])) <= 0) m--;
        ch[m++] = p[i];
    }
    return m;
}

Point vec[10], vec2[10];

int main() {
    scanf("%d", &n);
    for(int i = 0; i < n; i++)
        scanf("%lf%lf", &p[i].x, &p[i].y);
    cnt = ConvexHull(p, n, ch);

    cnt--;
    for(int i = 0; i < cnt; i++) {
        ch[cnt + i] = ch[i];
    }

    int pos1 = 1, pos2 = 1, pos3 = 1;
    double ans = inf;
    for(int i = 0; i < cnt; i++) {
        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])))
            pos1++;
        while(Dot(ch[i + 1] - ch[i], ch[pos2 + 1] - ch[i]) > Dot(ch[i + 1] - ch[i], ch[pos2] - ch[i]))
            pos2++;
        pos3 = max(pos3, pos1);
        while(Dot(ch[i + 1] - ch[i], ch[pos3 + 1] - ch[i]) < Dot(ch[i + 1] - ch[i], ch[pos3] - ch[i]))
            pos3++;
        Vector k1 = ch[i + 1] - ch[i];
        Vector k2 = Rotate(k1, PI / 2);
        Point p1 = GetLineIntersection(ch[i], k1, ch[pos2], k2);
        Point p2 = GetLineIntersection(ch[i], k1, ch[pos3], k2);
        Point p3 = GetLineIntersection(ch[pos1], k1, ch[pos2], k2);
        Point p4 = GetLineIntersection(ch[pos1], k1, ch[pos3], k2);
        double ret = dis(p1, p2) * dis(p1, p3);
        if(ret < ans) {
            ans = ret;
            vec[0] = p1;
            vec[1] = p2;
            vec[2] = p3;
            vec[3] = p4;
        }
    }

    ConvexHull(vec, 4, vec2);
    printf("%.5f\n", ans);
    for(int i = 0; i < 4; i++) {
        printf("%.5f %.5f\n", vec2[i].x, vec2[i].y);
    }
    return 0;
}
/*
*/