bzoj 4852 炸弹攻击
- 二维平面上的最优解问题,模拟退火是一个较为优秀的近似算法.
- 此题确定圆心后,便可 \(O(m)\) 算出收益,且最优解附近显然也较优,是连续变化的,可以直接模拟退火.
- 小技巧:这里不同答案差值比较小,而温度比较高,可以乘上一个常数,降低选取差解的概率,即满足下式时转移.出处.
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define mp make_pair
#define pii pair<int,int>
inline int read()
{
int x=0;
bool pos=1;
char ch=getchar();
for(;!isdigit(ch);ch=getchar())
if(ch=='-')
pos=0;
for(;isdigit(ch);ch=getchar())
x=x*10+ch-'0';
return pos?x:-x;
}
struct v2{
double x,y;
v2(double x=0,double y=0):x(x),y(y) {}
v2 operator + (const v2 &rhs) const
{
return v2(x+rhs.x,y+rhs.y);
}
v2 operator / (const double &rhs) const
{
return v2(x/rhs,y/rhs);
}
v2 operator - (const v2 &rhs) const
{
return v2(x-rhs.x,y-rhs.y);
}
double operator * (const v2 &rhs) const
{
return x*rhs.y-y*rhs.x;
}
double modulus()
{
return sqrt(x*x+y*y);
}
};
const int MAXN=1e3+10;
int n,m;
double R;
struct Tur{
double r;
v2 pos;
}tur[MAXN];
v2 enemy[MAXN];
int dcmp(double x)
{
return fabs(x)<=(1e-8)?0:(x>0?1:-1);
}
int calc(v2 centre)
{
double r=R;
for(int i=1;i<=n;++i)
r=min(r,(centre-tur[i].pos).modulus()-tur[i].r);
int res=0;
for(int i=1;i<=m;++i)
if(dcmp((centre-enemy[i]).modulus()-r)<=0)
++res;
return res;
}
double rd()
{
return (double) rand() / RAND_MAX;
}
double randpos(double &x,double &y)
{
x=2*R*rd()-R;
y=2*R*rd()-R;
}
const double dt=0.998;
const double eps=1e-2;
int cnt=0;
int SA(double x,double y)
{
int res=calc(v2(x,y));
int cur=res;
double T=R;
while(T>eps)
{
double nt=T+0.1;
double nx=x+ (2*nt)*rd()-nt,ny=y+(2*nt)*rd()-nt;
int nans=calc(v2(nx,ny));
if(nans>res || exp(1e4*(nans-cur)/T)>rd())
{
x=nx;
y=ny;
cur=nans;
}
res=max(res,cur);
T*=dt;
}
return res;
}
int main()
{
srand(19260817);
scanf("%d%d%lf",&n,&m,&R);
for(int i=1;i<=n;++i)
scanf("%lf%lf%lf",&tur[i].pos.x,&tur[i].pos.y,&tur[i].r);
for(int i=1;i<=m;++i)
scanf("%lf%lf",&enemy[i].x,&enemy[i].y);
int ans=0;
for(int i=1;i<=20;++i)
{
double x,y;
randpos(x,y);
ans=max(ans,SA(x,y));
}
printf("%d\n",ans);
return 0;
}