hdu 4766 Network

长春网络赛的题。

用的暴力。

XB看了题目，然后直接说他的思路，我俩讨论后敲的，我也没怎么看题。

我的做法是，先将额外的那个点（第一行输入M点），与每个hero为半径的圆求交点，若交点到各个hero的距离小于等于d，则最短是M到交点

所有hero的点以d为半径的圆求交，所交区域如果放上路由器一定能覆盖到所有的hero，所以变成M到这个区域的点的最短距离。

由于第一步的操作，所以M到点的最小距离的那个点一定是在相交圆弧的交点上，所以枚举交点，找最短距离。。。

复杂度N^3，太暴力了。

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define MID(x,y) ( ( x + y ) >> 1 )
#define L(x) ( x << 1 )
#define R(x) ( x << 1 | 1 )
#define FOR(i,s,t) for(int i=(s); i<(t); i++)
#define BUG puts("here!!!")
#define STOP system("pause")
#define file_r(x) freopen(x, "r", stdin)
#define file_w(x) freopen(x, "w", stdout)

using namespace std;

const int MAX = 1010;
const double eps = 1e-8;
const double INF = 1e30;
double d;
struct point{
	double x, y;
	void get()
	{
		scanf("%lf%lf", &x, &y);
	}
};
point M;
bool dy(double x,double y)	{	return x > y + eps;}	// x > y 
bool xy(double x,double y)	{	return x < y - eps;}	// x < y 
bool dyd(double x,double y)	{ 	return x > y - eps;}	// x >= y 
bool xyd(double x,double y)	{	return x < y + eps;} 	// x <= y 
bool dd(double x,double y) 	{	return fabs( x - y ) < eps;}  // x == y
double disp2p(point a,point b) //  a b 两点之间的距离 
{
	return sqrt( ( a.x - b.x ) * ( a.x - b.x ) + ( a.y - b.y ) * ( a.y - b.y ) );
}
bool operator==(point a, point b)
{
	return dd(a.x, b.x) && dd(a.y, b.y);
}
point c[MAX];

point l2l_inst_p(point u1,point u2,point v1,point v2)
{
	point ans = u1;
	double t = ((u1.x - v1.x)*(v1.y - v2.y) - (u1.y - v1.y)*(v1.x - v2.x))/
				((u1.x - u2.x)*(v1.y - v2.y) - (u1.y - u2.y)*(v1.x - v2.x));
	ans.x += (u2.x - u1.x)*t;
	ans.y += (u2.y - u1.y)*t;
	return ans;
}
void l2c_inst_p(point c,double r,point l1,point l2,point &p1,point &p2)
{
	point p = c;
	double t;
	p.x += l1.y - l2.y;
	p.y += l2.x - l1.x;
	p = l2l_inst_p(p,c,l1,l2);
	t = sqrt(r*r - disp2p(p,c)*disp2p(p,c))/disp2p(l1,l2);
	p1.x = p.x + (l2.x - l1.x)*t;
	p1.y = p.y + (l2.y - l1.y)*t;
	p2.x = p.x - (l2.x - l1.x)*t;
	p2.y = p.y - (l2.y - l1.y)*t;
}
void c2c_inst_p(point c1,double r1,point c2,double r2,point &p1,point &p2)
{
	point u,v;
	double t;
	t = (1 + (r1*r1 - r2*r2)/disp2p(c1,c2)/disp2p(c1,c2))/2;
	u.x = c1.x + (c2.x - c1.x)*t;
	u.y = c1.y + (c2.y - c1.y)*t;
	v.x = u.x + c1.y - c2.y;
	v.y = u.y - c1.x + c2.x;
	l2c_inst_p(c1,r1,u,v,p1,p2);
}
bool c2c_tangent(point a,double r1,point b,double r2)
{
	if( dd(disp2p(a,b),r1+r2) || dd(disp2p(a,b),fabs(r1-r2)) )
		return true;
	return false;
}
point c2c_tangent_p(point a,double r1,point b,double r2)
{
	point t;
	if( dd(disp2p(a,b),r1 + r2)  )
	{
		t.x = (r1*b.x + r2*a.x)/(r1 + r2);
		t.y = (r1*b.y + r2*a.y)/(r1 + r2);
		return t;
	}
	t.x = (r1*b.x - r2*a.x)/(r1 - r2);
	t.y = (r1*b.y - r2*a.y)/(r1 - r2);
	return t;
}
int fugai[MAX];
double crossProduct(point a,point b,point c)//向量 ac 在 ab 的方向 顺时针是正 
{
	return (c.x - a.x)*(b.y - a.y) - (b.x - a.x)*(c.y - a.y);
}
double disp2l(point a,point l1,point l2)
{
	return fabs( crossProduct(a,l1,l2) )/disp2p(l1,l2);
}
bool onSegment(point a, point b, point c)
{
	if( dd(crossProduct(a,b,c),0.0) && dyd(c.x,min(a.x,b.x)) && 
		xyd(c.x,max(a.x,b.x)) && dyd(c.y,min(a.y,b.y)) && xyd(c.y,max(a.y,b.y)) )
		return true;
	return false;
}

bool xiangqie(point p, int n) {
	FOR(i, 0, n)
		if( !dd(disp2p(p, c[i]), d) )
			return false;
	return true;
}
//M在相交区域内 
bool Minall(point M, int n) {
	FOR(i, 0, n)
		if( dy(disp2p(M, c[i]), d) )
			return false;
	return true;
}
double solve(int n) {
	if( Minall(M, n) )
		return 0.0;
	int cnt = 0;
	double mmin = INF;
	
	//若最后求的点在圆弧上 
	FOR(i, 0, n) {
		if( xyd(disp2p(M, c[i]), d) )
			continue;
		point a, b;
		l2c_inst_p(c[i], d, M, c[i], a, b);
		double len = disp2p(c[i], M);
		if( dd(len, disp2p(a, M) + d) && Minall(a, n) && xy(len-d, mmin) )
			mmin = len-d;
		if( dd(len, disp2p(b, M) + d) && Minall(b, n) && xy(len-d, mmin) )
			mmin = len-d;
	}
	if( xy(mmin, INF) )
		return mmin;
	FOR(i, 0, n)
		FOR(k, i+1, n) {
			if( dy(disp2p(c[i], c[k]), 2*d) )
				return -1;
			if( dd(disp2p(c[i], c[k]), 2*d) ) {
				point p;
				p.x = (c[i].x + c[k].x)/2;
				p.y = (c[i].y + c[k].y)/2;
				if( xiangqie(p, n) )
					return disp2p(p, M);
				else
					return -1;
			}
			if( c[i] == c[k] )
				continue;
			point a, b;
			c2c_inst_p(c[i], d, c[k], d, a, b);
			if( Minall(a, n) ) {
				double len = disp2p(M, a);
				if( xy(len, mmin) ) 
					mmin = len;
			}
			if( Minall(b, n) ) {
				double len = disp2p(M, b);
				if( xy(len, mmin) )
					mmin = len;
			}
		}
	return mmin;
}
int main()
{
	int ncases, n, ind = 1;
	while( ~scanf("%lf%lf%lf", &M.x, &M.y, &d) ) {
		scanf("%d", &n);
		FOR(i, 0, n)
			c[i].get();
		
		double ans = solve(n);
		if( ans < 0 )
			printf("X\n");
		else
			printf("%.2lf\n", ans);
	}

return 0;
}