POJ 3608 Bridge Across Islands (两凸包间最短距离 + 旋转卡壳)

题目：传送门

 

经典题

 

代码大部分参考了：kuangbin

 

 

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#include <map>
#include <vector>
#include <set>
#include <string>
#include <math.h>
#define LL long long
#define mem(i, j) memset(i, j, sizeof(i))
#define rep(i, j, k) for(int i = j; i <= k; i++)
#define dep(i, j, k) for(int i = k; i >= j; i--)
#define pb push_back
#define make make_pair
#define INF INT_MAX
#define inf LLONG_MAX
#define PI acos(-1)
using namespace std;

const int N = 5e5 + 5;

struct Point {
    double x, y;
    Point(double x = 0, double y = 0) : x(x), y(y) { } /// 构造函数
};

typedef Point Vector;
/// 向量+向量=向量， 点+向量=向量
Vector operator + (Vector A, Vector B) { return Vector(A.x + B.x, A.y + B.y); }
///点-点=向量
Vector operator - (Point A, Point B) { return Vector(A.x - B.x, A.y - B.y); }
///向量*数=向量
Vector operator * (Vector A, double p) { return Vector(A.x * p, A.y * p); }
///向量/数=向量
Vector operator / (Vector A, double p) { return Vector(A.x / p, A.y / p); }

const double eps = 1e-8;
int dcmp(double x) {
    if(fabs(x) < eps) return 0; else return x < 0 ? -1 : 1;
}

bool operator < (const Point& a, const Point& b) {
    return a.x == b.x ? a.y < b.y : a.x < b.x;
}

bool operator == (const Point& 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; } /// 点积
double Length(Vector A) { return sqrt(1.0 * Dot(A, A)); } /// 计算向量长度
double Angle(Vector A, Vector B) { return acos(Dot(A, B) / Length(A) / Length(B)); } /// 向量A、B夹角
double Cross(Vector A, Vector B) { return A.x*B.y - A.y*B.x; } /// 叉积

int ConvexHull(Point* p, int n, Point* ch) {
    sort(p, p + n);
    int m = 0;
    rep(i, 0, n - 1) {
        while(m > 1 && Cross(ch[m - 1] - ch[m - 2], p[i] - ch[m - 2]) <= 0) m--;
        ch[m++] = p[i];
    }
    int k = m;
    dep(i, 0, n - 2) {
        while(m > k && Cross(ch[m - 1] - ch[m - 2], p[i] - ch[m - 2]) <= 0) m--;
        ch[m++] = p[i];
    }
    if(n > 1) m--;
    return m;
}

double PTS(Point p, Point A, Point B) { /// 求点 p 到线段 AB 的最短距离
    if(A == B) return Length(p - A);
    Point v1 = B - A, v2 = p - A, v3 = p - B;
    if(dcmp(Dot(v1, v2)) < 0) return Length(v2);
    else if(dcmp(Dot(v1, v3)) > 0) return Length(v3);
    else return fabs(Cross(v1, v2)) / Length(v1);
}
/* 平行线段 a1a2 和 b1b2 的距离 */
double DS(Point a1, Point a2, Point b1, Point b2) {
    return min(PTS(b1, a1, a2), min(PTS(b2, a1, a2), min(PTS(a1, b1, b2), PTS(a2, b1, b2))));
}

/* 得到向量 a1a2 和 b1b2 的位置关系 */
double Get_angle(Point a1, Point a2, Point b1, Point b2) {
    Point t = b1 - (b2 - a1);
    return Cross(a2 - a1, t - a1);
}

double Rotating_calipers(Point P[], int n, Point Q[], int m) {
    int posp = 0, posq = 0;

    rep(i, 0, n - 1) if(dcmp(P[i].y - P[posp].y) < 0) posp = i;
    rep(i, 0, m - 1) if(dcmp(Q[i].y - Q[posq].y) > 0) posq = i;

    double tmp, ans = 1e99;

    rep(i, 0, n - 1) {
        while(dcmp(tmp = Cross(P[(posp + 1) % n] - P[posp], Q[posq] - P[posp]) - Cross(P[(posp + 1) % n] - P[posp], Q[(posq + 1) % m] - P[posp])) < 0)
            posq = (posq + 1) % m;
        if(dcmp(tmp) == 0)
            ans = min(ans, DS(P[posp], P[(posp + 1) % n], Q[posq], Q[(posq + 1) % m]));
        else
            ans = min(ans, PTS(Q[posq], P[posp], P[(posp + 1) % n]));
        posp = (posp + 1) % n;
    }

    return ans;
}

Point P[N], Q[N], p[N], q[N];

int main() {
    int n, m;
    while(scanf("%d %d", &n, &m) && n + m) {

        rep(i, 0, n - 1) scanf("%lf %lf", &p[i].x, &p[i].y);
        rep(i, 0, m - 1) scanf("%lf %lf", &q[i].x, &q[i].y);

        n = ConvexHull(p, n, P);
        m = ConvexHull(q, m, Q);

        double ans = min(Rotating_calipers(P, n, Q, m), Rotating_calipers(Q, m, P, n));

        printf("%.5f\n", ans);

    }
    return 0;
}