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模拟退火算法,125msAC具体算法参考《浅谈随机化思想在几何问题中的应用》。
C++ 125ms Accept!
void solve()
{
int p=10,l=25;
double delta,epsilon=1e-2,k1,k2,n1,n2,an[11];
delta=(double)max(x,y)/(sqrt(1.0*m));
for (int r=1;r<=p;r++)
{
q[r][0]=double(rand()%1000+1)/1000.000*x;
q[r][1]=double(rand()%1000+1)/1000.000*y;
double max=oo;
for (int i=1;i<=m;i++)
{
double tmp=dist(q[r][0],q[r][1],hole[i].xi,hole[i].yi);
if (tmp<max) max=tmp;
}
an[r]=max;
}
while (delta>epsilon)
{
for (int r=1;r<=p;r++)
{
for (int i=1;i<=l;i++)
{
k1=double(rand()%1000+1)/1000.000*delta;
k2=sqrt(delta*delta-k1*k1);
if (rand()%2==1) k1*=-1;
if (rand()%2==1) k2*=-1;
n1=q[r][0]+k1;
n2=q[r][1]+k2;
if (n1<=x&&n2<=y&&n1>=0&&n2>=0)
{
double max1=oo;
for (int j=1;j<=m;j++)
{
double tmp=dist(n1,n2,hole[j].xi,hole[j].yi);
if (tmp<max1) max1=tmp;
}
if (max1>an[r])
q[r][0]=n1,q[r][1]=n2,an[r]=max1;
}
else
continue;
}
}
delta*=0.8;
}
double max=-oo;
int s1=0;
for (int r=1;r<=p;r++)
if (an[r]>max)
max=an[r],s1=r;
ans.xi=q[s1][0];
ans.yi=q[s1][1];
}
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