POJ 1379 Run Away (Basic simulated annealing)

Question: find out one point. The shortest distance from all points is the maximum.
 
Http://poj.org/problem? Id = 1379
 
 
Randomization step size, greedy adjustment.
 
What we need to prove with Simulated Annealing ~~~~ % # @ & % ...... @ % ￥ % @ Proof complete
 
I have nothing to say. The time limit and accuracy requirements are not high. Just do it.
 
 
[Cpp]
# Include <iostream>
# Include <cstdio>
# Include <cstring>
# Include <queue>
# Include <cmath>
# Include <string>
# Include <vector>
# Include <algorithm>
# Include <map>
# Include <set>
# Include <ctime>
# Define maxn200005
# Define eps 1e-8
# Define inf 2000000000
# Define LL long
# Define zero (a) fabs (a) <eps
# Define MOD 19901014.
# Define N 1000005
# Define pi acos (-1.0)
Using namespace std;
Int t, n;
Double X, Y;
Double best [50];
Struct Point {
Double x, y;
Bool check (){
If (x>-eps & x <eps + X & y>-eps & y <eps + Y)
Return true;
Return false;
}
} P [2, 1005], tp [50];
Double dist (Point p1, Point p2 ){
Return sqrt (p1.x-p2.x) * (p1.x-p2.x) + (p1.y-p2.y) * (p1.y-p2.y ));
}
Double slove (Point p0 ){
Double ans = inf;
For (int I = 0; I <n; I ++)
Ans = min (ans, dist (p [I], p0 ));
Return ans;
}
Int main (){
Scanf ("% d", & t );
Srand (time (NULL ));
While (t --){
Scanf ("% lf % d", & X, & Y, & n );
For (int I = 0; I <n; I ++)
Scanf ("% lf", & p [I]. x, & p [I]. y );
For (int I = 0; I <15; I ++ ){
Tp [I]. x = (rand () % 1000 + 1)/1000.0 * X;
Tp [I]. y = (rand () % 1000 + 1)/1000.0 * Y;
Best [I] = slove (tp [I]);
}
Double step = max (X, Y)/sqrt (1.0 * n );
While (step> 1e-3 ){
For (int I = 0; I <15; I ++ ){
Point cur, pre = tp [I];
For (int j = 0; j <35; j ++ ){
Double angle = (rand () % 1000 + 1)/1000.0*2 * pi;
Cur. x = pre. x + cos (angle) * step;
Cur. y = pre. y + sin (angle) * step;
If (! Cur. check () continue;
Double tmp = slove (cur );
If (tmp> best [I]) {
Tp [I] = cur;
Best [I] = tmp;
}
}
}
Step * = 0.85;
}
Int idx = 0;
Double ans = 0;
For (int I = 0; I <15; I ++ ){
If (best [I]> ans ){
Ans = best [I];
Idx = I;
}
}
Printf ("The safest point is (%. 1f, %. 1f). \ n", tp [idx]. x, tp [idx]. y );
}
Return 0;
}
 
# Include <iostream>
# Include <cstdio>
# Include <cstring>
# Include <queue>
# Include <cmath>
# Include <string>
# Include <vector>
# Include <algorithm>
# Include <map>
# Include <set>
# Include <ctime>
# Define maxn200005
# Define eps 1e-8
# Define inf 2000000000
# Define LL long
# Define zero (a) fabs (a) <eps
# Define MOD 19901014.
# Define N 1000005
# Define pi acos (-1.0)
Using namespace std;
Int t, n;
Double X, Y;
Double best [50];
Struct Point {
Double x, y;
Bool check (){
If (x>-eps & x <eps + X & y>-eps & y <eps + Y)
Return true;
Return false;
}
} P [2, 1005], tp [50];
Double dist (Point p1, Point p2 ){
Return sqrt (p1.x-p2.x) * (p1.x-p2.x) + (p1.y-p2.y) * (p1.y-p2.y ));
}
Double slove (Point p0 ){
Double ans = inf;
For (int I = 0; I <n; I ++)
Ans = min (ans, dist (p [I], p0 ));
Return ans;
}
Int main (){
Scanf ("% d", & t );
Srand (time (NULL ));
While (t --){
Scanf ("% lf % d", & X, & Y, & n );
For (int I = 0; I <n; I ++)
Scanf ("% lf", & p [I]. x, & p [I]. y );
For (int I = 0; I <15; I ++ ){
Tp [I]. x = (rand () % 1000 + 1)/1000.0 * X;
Tp [I]. y = (rand () % 1000 + 1)/1000.0 * Y;
Best [I] = slove (tp [I]);
}
Double step = max (X, Y)/sqrt (1.0 * n );
While (step> 1e-3 ){
For (int I = 0; I <15; I ++ ){
Point cur, pre = tp [I];
For (int j = 0; j <35; j ++ ){
Double angle = (rand () % 1000 + 1)/1000.0*2 * pi;
Cur. x = pre. x + cos (angle) * step;
Cur. y = pre. y + sin (angle) * step;
If (! Cur. check () continue;
Double tmp = slove (cur );
If (tmp> best [I]) {
Tp [I] = cur;
Best [I] = tmp;
}
}
}
Step * = 0.85;
}
Int idx = 0;
Double ans = 0;
For (int I = 0; I <15; I ++ ){
If (best [I]> ans ){
Ans = best [I];
Idx = I;
}
}
Printf ("The safest point is (%. 1f, %. 1f). \ n", tp [idx]. x, tp [idx]. y );
}
Return 0;
}