POJ 1379 Run Away 【基础模拟退火】

题意：找出一点，距离所有所有点的最短距离最大

二维平面内模拟退火即可，同样这题用最小圆覆盖也是可以的。

 

Source Code:

//#pragma comment(linker, "/STACK:16777216") //for c++ Compiler

#include <stdio.h>

#include <iostream>

#include <fstream>

#include <cstring>

#include <cmath>

#include <stack>

#include <string>

#include <map>

#include <queue>

#include <vector>

#include <ctime>

#include <algorithm>

#define LL long long

#define Max(a,b) (((a) > (b)) ? (a) : (b))

#define Min(a,b) (((a) < (b)) ? (a) : (b))

#define Abs(x) (((x) > 0) ? (x) : (-(x)))

#define MOD 1000000007

#define eps 1e-8

#define pi acos(-1.0)



using namespace std;



const int inf = 0x3f3f3f3f;

const int N = 15;

const int L = 35;



int t,n;

double X ,Y, 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[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 min_dis(Point p0){

    double ans = inf;//

    for(int i = 0; i < n; ++i)

        ans = min(ans,dist(p[i],p0));//

    return ans;

}



Point rand_point(double x, double y){

    Point c;

    c.x = (rand() % 1000 + 1) / 1000.0 * x;

    c.y = (rand() % 1000 + 1) / 1000.0 * y;

    return c;

}



int main(){

    srand(time(NULL));

    scanf("%d",&t);

    while(t--){

        scanf("%lf%lf%d",&X,&Y,&n);

        for(int i = 0; i < n; ++i)

            scanf("%lf%lf",&p[i].x,&p[i].y);

        for(int i = 0; i < N; ++i){

            tp[i] = rand_point(X, Y);

            best[i] = min_dis(tp[i]);

        }

        double step = max(X,Y) / sqrt(1.0 * n);

        while(step > 1e-3){

            for(int i = 0; i < N; ++i){

                Point cur;

                Point pre = tp[i];

                for(int j = 0; j < L; ++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 = min_dis(cur);

                    if(tmp > best[i]){//

                        tp[i] = cur;

                        best[i] = tmp;

                    }

                }

            }

            step *= 0.85;

        }

        int idx = 0;

        for(int i = 0; i < N; ++i){

            if(best[i] > best[idx]){//

                idx = i;

            }

        }

        printf("The safest point is (%.1f, %.1f).\n",tp[idx].x,tp[idx].y);

        //printf("%.1f\n",best[idx]);

    }

    return 0;

}