poj 1379 Run Away 模拟退火

Run Away

Time Limit: 3000MS Memory Limit: 65536KTotal Submissions: 7520 Accepted: 2280

Description

One of the traps we will encounter in the Pyramid is located in the Large Room. A lot of small holes are drilled into the floor. They look completely harmless at the first sight. But when activated, they start to throw out very hot java, uh ... pardon, lava. Unfortunately, all known paths to the Center Room (where the Sarcophagus is) contain a trigger that activates the trap. The ACM were not able to avoid that. But they have carefully monitored the positions of all the holes. So it is important to find the place in the Large Room that has the maximal distance from all the holes. This place is the safest in the entire room and the archaeologist has to hide there.

Input

The input consists of T test cases. The number of them (T) is given on the first line of the input file. Each test case begins with a line containing three integers X, Y, M separated by space. The numbers satisfy conditions: 1 <= X,Y <=10000, 1 <= M <= 1000. The numbers X and Yindicate the dimensions of the Large Room which has a rectangular shape. The number M stands for the number of holes. Then exactly M lines follow, each containing two integer numbers Ui and Vi (0 <= Ui <= X, 0 <= Vi <= Y) indicating the coordinates of one hole. There may be several holes at the same position.

Output

Print exactly one line for each test case. The line should contain the sentence "The safest point is (P, Q)." where P and Qare the coordinates of the point in the room that has the maximum distance from the nearest hole, rounded to the nearest number with exactly one digit after the decimal point (0.05 rounds up to 0.1).

Sample Input

31000 50 110 10100 100 410 1010 9090 1090 903000 3000 41200 8563 25002700 2650 2990 100

Sample Output

The safest point is (1000.0, 50.0).The safest point is (50.0, 50.0).The safest point is (1433.0, 1669.8).

题意：在一个范围内给出一些点，找出一个点，使得这个点到它最近点的距离最大。

思路：模拟退火。

代码：

#include <cstdio>#include <cmath>#include <ctime>#include <cstdlib>#include <algorithm>using namespace std;#define eps 1e-6#define maxn 1000+10#define inf 1e20#define PI acos(-1.0)double X,Y;double x[maxn],y[maxn];double px[maxn],py[maxn],d[maxn];int n;double dist(double x1,double y1,double x2,double y2){return sqrt((x1-x2)*(x1-x2) + (y1-y2)*(y1-y2));}int main(){int T;scanf("%d",&T);//srand(time(NULL));while(T--){scanf("%lf %lf %d",&X,&Y,&n);for(int i=0;i<n;i++){scanf("%lf %lf",&x[i],&y[i]);}px[0] = 0.0, py[0] = 0.0;px[1] = 0.0, py[1] = Y;px[2] = X, py[2] = 0.0;px[3] = X, py[3] = Y;  for(int i=0;i<30;i++){if(i>=4){px[i] = (rand()%10005+1)/10005.0*X;py[i] = (rand()%10005+1)/10005.0*Y;}double t = inf;for(int j=0;j<n;j++){t = min(t,dist(px[i],py[i],x[j],y[j]));}d[i] = t;}double delta = max(X,Y)/sqrt(1.0*n);while(delta>eps){for(int i=0;i<30;i++){for(int j=0;j<10;j++){double a = (rand()%10005+1)/10005.0*2*PI;double dx = px[i] + delta*cos(a);double dy = py[i] + delta*sin(a);if(dx<0 || dx>X || dy<0 || dy>Y) continue;double t = inf;for(int k=0;k<n;k++){t = min(t,dist(dx,dy,x[k],y[k]));}if(t>d[i]) {d[i] = t;px[i] = dx;py[i] = dy;}}}delta *= 0.85;}double ans = 0.0;int q = 0;for(int i=0;i<30;i++){if(d[i]>ans){ans = d[i];q = i;}}printf("The safest point is (%.1f, %.1f).\n",px[q],py[q]);}return 0;}