BZOJ1844/2210Pku1379 Run Away 模拟退火

Posted 2020-09-30 CQzhangyu

【BZOJ1844/2210】Pku1379 Run Away

题意：矩形区域中有一堆点，求矩形中一个位置使得它到所有点的距离的最小值最大。

题解：模拟退火的裸题，再调调调调调参就行了~

#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
#include <cstdlib>
using namespace std;
const int maxn=1010;
struct pdd
{
	double x,y;
	pdd() {}
	pdd(double a,double b) {x=a,y=b;}
}p[maxn];
int n;
double X,Y,T,mx;
pdd ans,now,neo;
double getdis(pdd a,pdd b)
{
	return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
int rd()
{
	int ret=0,f=1;	char gc=getchar();
	while(gc<‘0‘||gc>‘9‘)	{if(gc==‘-‘)f=-f;	gc=getchar();}
	while(gc>=‘0‘&&gc<=‘9‘)	ret=ret*10+gc-‘0‘,gc=getchar();
	return ret*f;
}
double solve(pdd a)
{
	double ret=1e10;
	for(int i=1;i<=n;i++)	ret=min(ret,getdis(a,p[i]));
	if(ret>mx)	mx=ret,ans=a;
	return ret;
}
double Rand()
{
	return rand()%1000/1000.0;
}
void work()
{
	X=rd(),Y=rd(),n=rd();
	int i,j;
	for(i=1;i<=n;i++)	p[i].x=rd(),p[i].y=rd();
	mx=0;
	for(j=1;j<=50;j++)
	{
		now.x=Rand()*X,now.y=Rand()*Y,solve(now),T=1000;
		while(T>1e-3)
		{
			double a=2.0*acos(-1.0)*Rand();
			neo.x=now.x+T*cos(a),neo.y=now.y+T*sin(a),T*=0.95;
			if(neo.x<0||neo.x>X||neo.y<0||neo.y>Y)	continue;
			double de=solve(neo)-solve(now);
			if(de>0)	now=neo;
		}
		T=0.5;
		for(i=1;i<=500;i++)
		{
			double a=2.0*acos(-1.0)*Rand();
			neo.x=now.x+T*cos(a);
			neo.y=now.y+T*sin(a);
			if(neo.x<0||neo.x>X||neo.y<0||neo.y>Y)	continue;
			solve(neo);
		}
	}
	printf("The safest point is (%.1lf, %.1lf).\n",ans.x,ans.y);
}
int main()
{
	srand(2333666);
	int T=rd();
	while(T--)	work();
}