HDU 4667 Building Fence 计算几何 凸包+圆

1.三角形的所有端点

2.过所有三角形的端点对所有圆做切线，得到所有切点。

3.做任意两圆的外公切线，得到所有切点。

对上述所有点求凸包，标记每个点是三角形上的点还是某个圆上的点。

求完凸包后，因为所有点都是按逆时针（或顺时针）排好序的，如果相邻两点在同一圆上，那么求这段圆弧的距离，否则求这段直线的距离。最后得到所有周长。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>

using namespace std;

const double eps = 1e-9;
const double PI = acos(-1.0);
const int MAXN = 60;

struct Point
{
double x, y;
int id;         //点标号，标记是否在同一个圆上
Point() { }
Point( double x, double y ):x(x), y(y) { }
Point( double x, double y, int id ):x(x), y(y), id(id) { }
void readPoint()
{
scanf( "%lf%lf", &x, &y );
return;
}
};

struct Circle
{
Point c;   //圆心坐标
double r;  //半径
Circle() {}
Circle( Point c, double r ): c(c), r(r) {}
Point getPoint( double theta )   //根据极角返回圆上一点的坐标
{
return Point( c.x + cos(theta)*r, c.y + sin(theta)*r );
}
void readCircle()
{
scanf("%lf%lf%lf", &c.x, &c.y, &r );
return;
}
};

typedef Point Vector;

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

Vector operator-( Vector A, Vector B )       //向量减
{
return Vector( A.x - B.x, A.y - B.y );
}

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

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

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

bool operator<( const Point& A, const Point& B )   //两点比较
{
return dcmp( A.x - B.x) < 0 || ( dcmp(A.x - B.x ) == 0 && dcmp( A.y - B.y ) < 0 );
}

bool operator>( const Point& A, const Point& B )   //两点比较
{
return dcmp( A.x - B.x) > 0 || ( dcmp(A.x - B.x ) == 0 && dcmp( A.y - B.y ) > 0 );
}

bool operator==( const Point& a, const Point& b )   //两点相等
{
return dcmp( a.x - b.x ) == 0 && dcmp( a.y - b.y ) == 0;
}

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

double PointDis( Point a, Point b ) //两点距离的平方
{
return (a.x - b.x)*(a.x - b.x) + (a.y - b.y)*(a.y - b.y);
}

//求凸包，graham算法，O(nlogn)，返回凸包点的个数
int graham( Point *p, int n, Point *ch )
{
if ( n <= 2 ) return 0;
int top = 0;
sort( p, p + n );

ch[ top ] = p[0];
ch[ ++top ] = p[1];
ch[ ++top ] = p[2];

top = 1;

for ( int i = 2; i < n; ++i )
{
while ( top && dcmp( Cross( ch[top] - ch[top - 1], p[i] - ch[top - 1] ) ) <= 0 ) --top;
ch[++top] = p[i];
}
int len = top;
ch[++top] = p[n - 2];
for ( int i = n - 3; i >= 0; --i )
{
while ( top > len && dcmp( Cross( ch[top] - ch[top - 1], p[i] - ch[top - 1] ) ) <= 0 ) --top;
ch[++top] = p[i];
}
return top;
}

//过定点做圆的切线,得到切点，返回切点个数
//tps保存切点坐标
int getTangentPoints( Point p, Circle C, Point *tps )
{
int cnt = 0;

double dis = sqrt( PointDis( p, C.c ) );
int aa = dcmp( dis - C.r );
if ( aa < 0 ) return 0;  //点在圆内
else if ( aa == 0 ) //点在圆上，该点就是切点
{
tps[cnt] = p;
++cnt;
return cnt;
}

//点在圆外，有两个切点
double base = atan2( p.y - C.c.y, p.x - C.c.x );
double ang = acos( C.r / dis );
//printf( "base = %f ang=%f\n", base, ang );
//printf( "base-ang=%f  base+ang=%f \n", base - ang, base + ang );

tps[cnt] = C.getPoint( base - ang ), ++cnt;
tps[cnt] = C.getPoint( base + ang ), ++cnt;

return cnt;
}

//求两圆外公切线切点，返回切线个数
//p是圆c2在圆c1上的切点
int makeCircle( Circle c1, Circle c2, Point *p )
{
int cnt = 0;
double d = sqrt( PointDis(c1.c, c2.c) ), dr = c1.r - c2.r;
double b = acos(dr / d);
double a = atan2( c2.c.y - c1.c.y, c2.c.x - c1.c.x );
double a1 = a - b, a2 = a + b;
p[cnt++] = Point(cos(a1) * c1.r, sin(a1) * c1.r) + c1.c;
p[cnt++] = Point(cos(a2) * c1.r, sin(a2) * c1.r) + c1.c;
return cnt;
}

double DisOnCircle( Point a, Point b, Circle c )  //求圆上一段弧长
{
Point o = c.c;
double A = sqrt( PointDis( o, a ) );
double B = sqrt( PointDis( o, b ) );
double C = sqrt( PointDis( a, b ) );
double alpha = acos( ( A*A + B*B - C*C ) / ( 2.0*A*B ) );
if ( dcmp( Cross( o-a, o-b ) ) < 0 ) return alpha * c.r;
else return ( 2.0*PI - alpha ) * c.r;
}

/**********************以上模板**********************/

int cntC, cntT;  //圆的个数，三角形的个数
Circle yuan[MAXN];   //所有圆
Point PP[300100];    //所有点
Point tubao[300100]; //凸包
int totPP;           //点总数

void showP( Point *p, int nn )
{
printf( "allP = %d\n", nn );
for ( int i = 0; i < nn; ++i )
printf("%f %f\n", p[i].x, p[i].y );
puts("-------------------------");
return;
}

int main()
{
//freopen( "10022.in", "r", stdin );
//freopen( "s.out", "w", stdout );
while ( scanf( "%d%d", &cntC, &cntT ) == 2 )
{
totPP = 0;
for ( int i = 0; i < cntC; ++i )
yuan[i].readCircle();
for ( int i = 0; i < cntT; ++i )
{
for ( int j = 0; j < 3; ++j )
{
PP[totPP].readPoint();
PP[totPP].id = -(totPP+2);
++totPP;
}
}

if ( cntC == 1 && cntT == 0 )
{
printf("%.6f\n", 2.0 * PI * yuan[0].r );
continue;
}

int pretot = totPP;
//求两圆的外切点
for ( int i = 0; i < cntC; ++i )
for ( int j = i + 1; j < cntC; ++j )
{
Point PonA[4], PonB[4];
makeCircle( yuan[i], yuan[j], PonA );
int ans = makeCircle( yuan[j], yuan[i], PonB );
for ( int k = 0; k < ans; ++k )
{
PonA[k].id = i;
PonB[k].id = j;
PP[totPP++] = PonA[k];
PP[totPP++] = PonB[k];
}
}

//求所有点与所有圆的切点
for ( int i = 0; i < pretot; ++i )
{
for ( int j = 0; j < cntC; ++j )
{
Point qiedian[4];
int ans = getTangentPoints( PP[i], yuan[j], qiedian );
for ( int k = 0; k < ans; ++k )
{
qiedian[k].id = j;
PP[totPP++] = qiedian[k];
}
}
}

//showP( PP, totPP );
int cntBao = graham( PP, totPP, tubao );
//puts("*********");
//showP( tubao, cntBao );
double girth = 0.0;
tubao[cntBao] = tubao[0];

for ( int i = 1; i <= cntBao; ++i )
{
if ( tubao[i].id == tubao[i - 1].id )  //如果两点在同一个圆上
girth += DisOnCircle( tubao[i], tubao[i - 1], yuan[ tubao[i].id ] );
else
girth += sqrt( PointDis( tubao[i], tubao[i - 1] ) );

}

printf( "%.5lf\n", girth );
}
return 0;
}