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).
",tp[idx].x,tp[idx].y);
        //printf("%.1f
",best[idx]);
    }
    return 0;
}