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poj 2187 凸包or旋转qia壳法

2015-07-29 19:26 330 查看
题意:

给n(50000)个点,求这些点与点之间距离最大的距离。

解析:

先求凸包然后暴力。

或者旋转卡壳大法。

代码:

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <stack>
#include <vector>
#include <queue>
#include <map>
#include <climits>
#include <cassert>
#define LL long long

using namespace std;
const int inf = 0x3f3f3f3f;
const double eps = 1e-8;
const double pi = acos(-1.0);
const double ee = exp(1.0);

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

bool cmp(Point A, Point B)
{
if (A.x == B.x)
return A.y < B.y;
return A.x < B.x;
}

typedef Point Vector;

Vector operator + (Vector A, Vector B)
{
return Vector(A.x + B.x, A.y + B.y);
}

Vector operator - (Point A, Point B)
{
return Vector(A.x - B.x, A.y - B.y);
}

Vector operator * (Vector A, double p)
{
return Vector(A.x * p, A.y * p);
}

Vector operator / (Vector A, double p)
{
return Vector(A.x / p, A.y / p);
}

//便于点排序
bool operator < (const Point& a, const Point& b)
{
return a.x < b.x || (a.x == b.x && a.y < b.y);
}

//用于判断相等
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 dcmp(a.x - b.x) == 0 && dcmp(a.y - b.y) == 0;
}

//点积
double Dot(Vector A, Vector B)
{
return A.x * B.x + A.y * B.y;
}

//向量长度
double Length(Vector A)
{
return sqrt(Dot(A, A));
}

//向量夹角
double Angle(Vector A, Vector B)
{
return acos(Dot(A, B) / Length(A) / Length(B));
}

//叉集
double Cross(Vector A, Vector B)
{
return A.x * B.y - A.y * B.x;
}

//求两个向量相夹的面积
double Area2(Point A, Point B, Point C)
{
return Cross(B - A, C - A);
}

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

//向量的单位法线即向量左转90°
Vector Normal(Vector A)
{
double L = Length(A);
return Vector(-A.y / L, A.x / L);
}

//求交点坐标
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;
}

//点P到直线AB的距离
double DistanceToLine(Point P, Point A, Point B)
{
Vector v1 = B - A;
Vector v2 = P - A;
return fabs(Cross(v1, v2)) / Length(v1);
}

//点到直线的距离
double DistanceToSegment(Point P, Point A, Point B)
{
if (A == B)
{
return Length(P - A);
}
Vector v1 = B - A;
Vector v2 = P - A;
Vector 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);
}

//判断线段是否相交
bool SegmentProperIntersection(Point a1, Point a2, Point b1, Point b2)
{
double c1 = Cross(a2 - a1, b1 - a1), c2 = Cross(a2 - a1, b2 - a1);
double c3 = Cross(b2 - b1, a1 - b1), c4 = Cross(b2 - b1, a2 - b1);
return dcmp(c1) * dcmp(c2) < 0 && dcmp(c3) * dcmp(c4) < 0;
}

//线段是否在端点相交 + 线段相交的判定
bool OnSegment(Point p, Point a1, Point a2)
{
return dcmp(Cross(a1 - p, a2 - p)) == 0 && dcmp(Dot(a1 - p, a2 - p)) < 0;
}

//传入顶点集 计算多边形的面积
double ConvexPolygonArea(Point* p, int n)
{
double area = 0;
for (int i = 1; i < n - 1; i++)
{
area += Cross(p[i] - p[0], p[i + 1] - p[0]);
}
return area / 2.0;
}

//° -> 弧度
double torad(double deg)
{
return deg / 180 * pi;
}

//求凸包 返回点个数 ch为凸包的点
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 && 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 && Cross(ch[m - 1] - ch[m - 2], p[i] - ch[m - 2]) <= 0)
m--;
ch[m++] = p[i];
}
if (n > 1)
m--;
return m;
}

Point readPoint()
{
double x, y;
scanf("%lf %lf", &x, &y);
return Point(x, y);
}

//////////////////////////////////////////////////////
const int maxn = 50000 + 10;

int n;
Point p[maxn << 1];
Point ch[maxn << 1];

double dist(Point a, Point b)
{
return Dot(a - b, a - b);
}

void bruteForce()
{
double ans = 0;
int m = ConvexHull(p, n, ch);
for (int i = 0; i < m; i++)
{
for (int j = i + 1; j < m; j++)
{
ans = max(ans, dist(ch[i], ch[j]));
}
}
printf("%.0lf\n", ans);
}

void rotatingCalipers()
{
double ans = 0;
int m = ConvexHull(p, n, ch);
if (m == 2)
{
printf("%.0lf\n", dist(ch[0], ch[1]));
return;
}
int i = 0, j = 0;
for (int k = 0; k < m; k++)
{
if (ch[k] < ch[i])
i = k;
if (ch[j] < ch[k])
j = k;
}
int si = i, sj = j;
while (i != sj || j != si)
{
ans = max(ans, dist(ch[i], ch[j]));
if (Cross(ch[(i + 1) % m] - ch[i], ch[(j + 1) % m] - ch[j]) < 0)
{
i = (i + 1) % m;
}
else
{
j = (j + 1) % m;
}
}
printf("%.0lf\n", ans);
}

int main()
{
#ifdef LOCAL
freopen("in.txt", "r", stdin);
#endif // LOCAL
while (~scanf("%d", &n))
{
for (int i = 0; i < n; i++)
{
p[i] = readPoint();
}
rotatingCalipers();
}
return 0;
}
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