HDU 3932 模拟退火

HDU3932

 题目大意：给定一堆点，找到一个点的位置使这个点到所有点中的最大距离最小

简单的模拟退火即可

 

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <cmath>
 6 #include <ctime>
 7 #include <algorithm>
 8 
 9 using namespace std;
10 
11 #define N 1005
12 #define PI acos(-1.0)
13 #define random(x) (rand()%x+1)
14 const int P = 20;
15 const int L = 25;
16 double X,Y;
17 int n;
18 double mindis[N];
19 
20 struct Point{
21     double x , y;
22     Point(double x=0 , double y=0):x(x),y(y){}
23     void input(){
24         scanf("%lf%lf" , &x , &y);
25     }
26 }p[N] , tmp[N];
27 
28 double dis(Point a , Point b)
29 {
30     double x = a.x-b.x , y=a.y-b.y;
31     return sqrt(x*x+y*y);
32 }
33 
34 double cal(Point a)
35 {
36     double maxn = 0;
37     for(int i=0 ; i<n ; i++) maxn = max(maxn , dis(a , p[i]));
38     return maxn;
39 }
40 
41 int main()
42 {
43     #ifndef ONLINE_JUDGE
44         freopen("a.in" , "r" , stdin);
45     #endif // ONLINE_JUDGE
46     while(~scanf("%lf%lf%d" , &X , &Y , &n))
47     {
48         for(int i=0 ; i<n ; i++) p[i].input();
49         for(int i=0 ; i<P ; i++){
50             tmp[i].x = random(1000)/1000.0*X;
51             tmp[i].y = random(1000)/1000.0*Y;
52             mindis[i] = cal(tmp[i]);
53         }
54         double step = sqrt(X*X+Y*Y)/2;
55         while(step>1e-3){
56             for(int i=0 ; i<P ; i++){
57                 for(int j=0 ; j<L ; j++){
58                     Point cur;
59                     double ang = random(1000)/1000.0*2*PI;
60                     cur.x = tmp[i].x+cos(ang)*step;
61                     cur.y = tmp[i].y+sin(ang)*step;
62                     if(cur.x<0 || cur.x>X || cur.y<0 || cur.y>Y) continue;
63                     double val = cal(cur);
64                     if(val<mindis[i]){
65                         mindis[i] = val;
66                         tmp[i] = cur;
67                     }
68                 }
69             }
70             step *= 0.85;
71         }
72         double ret = 1e20;
73         Point u;
74         for(int i=0 ; i<P ; i++){
75             if(mindis[i]<ret){
76                 u = tmp[i];
77                 ret = mindis[i];
78             }
79         }
80         printf("(%.1f,%.1f).\n%.1f\n" , u.x,u.y,ret);
81     }
82     return 0;
83 }

 

 posted on 2015-05-26 14:44  Love风吟  阅读(395)  评论(0)  编辑  收藏  举报