目录

题目链接

loj#2076. 「JSOI2016」炸弹攻击

题解

模拟退火

退火时,由于答案比较小,但是温度比较高

所以在算exp时最好把相差的点数乘以一个常数让选取更差的的概率降低

代码

#include<ctime>
#include<cmath>
#include<cstdio>
#include<cstring>
#include<algorithm>
#define gc getchar()
#define pc putchar
inline int read() {
int x = 0,f = 1;
char c = gc;
while(c < '0' || c > '9') c =gc;
while(c <= '9' && c >= '0') x = x * 10 + c - '0',c = gc;
return x * f;
}
void print(int x) {
if(x < 0) {
pc('-');
x = -x;
}
if(x >= 10) print(x / 10);
pc(x % 10 + '0');
}
const int maxn = 100007;
int n,m; double R;
double X[maxn],Y[maxn],r[maxn],p[maxn],q[maxn];
double rd() {
return (double) rand() / RAND_MAX;
}
void randpos(double &x,double & y){
x = 2 * R * rd() - R;
y = 2 * R * rd() - R;
}
#define dt 0.998
#define eps 1e-2
double dis(double x,double y,double x1,double y1) {
return std::sqrt((x - x1) * (x - x1) + (y - y1) * (y - y1));
}
int calc(double x,double y) {
int ret = 0;
double tr = R;
for(int i = 1;i <= n;++ i) {
tr = std::min(tr,dis(X[i],Y[i],x,y) - r[i]);
}
if(r < 0) return 0;
for(int i = 1;i <= m;++ i)
if(dis(x,y,p[i],q[i]) <= tr) ret ++;
return ret;
}
int solve(double x,double y) {
int mx = calc(x,y);
int now = mx;
double T = R;
for(;T > eps;T *= dt) {
double nt = T + 0.1;
double nx = x + (2.0 * nt) * rd() - nt,ny = y + (2 * nt) * rd() - nt;
int nans = calc(nx,ny);
if(nans > mx || exp(1e4 * (nans - now) / T) > rd()) x = nx,y = ny,now = nans;
mx = std::max(mx,now);
}
return mx;
}
int main() {
srand(19991206);
scanf("%d%d%lf",&n,&m,&R);
for(int i = 1;i <= n;++ i)
scanf("%lf%lf%lf",&X[i],&Y[i],&r[i]);
for(int i = 1;i <= m;++ i)
scanf("%lf%lf",p + i,q + i);
int ans = 0 ;
double px,py;
for(int i = 1;i <= 20;++ i) {
randpos(px,py);
ans = std::max(ans,solve(px,py));
}
print(ans);
pc('\n');
}
05-06 10:42