poj1379 Run Away

Posted 2020-09-18 逢山开路 遇水架桥

传送门：http://poj.org/problem?id=1379

【题解】

题目大意：求(0,0)->(X,Y)内的一个点，使得这个点到给定的n个点的最小距离最大。

模拟退火

一开始可以先把4个顶点加入。

调调参就过样例了。

然后就过了

# include <math.h> 
# include <stdio.h>
# include <stdlib.h>
# include <string.h>
# include <iostream>
# include <algorithm>
// # include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int M = 5e5 + 10;
const int mod = 1e9+7;
const double pi = acos(-1.0);

# define RG register
# define ST static

double X, Y;
int n;
struct pa {
    double x, y;
    double dis;
    pa() {}
    pa(double x, double y, double dis) : x(x), y(y), dis(dis) {}
}a[M];


namespace SA {
    const double eps = 1e-2, DEC = 0.9, ACCEPT_DEC = 0.5;
    const int N = 30, T = 30, RAD = 1000;
    inline double rand01() {
        return rand() % (RAD + 1) / (1.0 * RAD);
    }
    inline double getdist(double x, double y) {
        double ret = 1e18;
        for (int i=1; i<=n; ++i) 
            ret = min(ret, (x-a[i].x)*(x-a[i].x)+(y-a[i].y)*(y-a[i].y));
        return ret;
    }
    pa res[N + 5];
    inline pa main() {
        res[1] = pa(0, 0, getdist(0, 0));
        res[2] = pa(X, 0, getdist(X, 0));
        res[3] = pa(0, Y, getdist(0, Y));
        res[4] = pa(X, Y, getdist(X, Y));
        for (int i=5; i<=N; ++i) {
            double x = rand01() * X;
            double y = rand01() * Y;
            res[i] = pa(x, y, getdist(x, y));
        }
        double temper = max(X, Y), accept = 0.6;
        while(temper > eps) {
            for (int i=1; i<=N; ++i) {
                for (int j=1; j<=T; ++j) {
                    double theta = rand01() * 2 * pi;
                    double xx = res[i].x + temper * cos(theta);
                    double yy = res[i].y + temper * sin(theta);
                    if(0 <= xx && xx <= X && 0 <= yy && yy <= Y) {
                        pa t = pa(xx, yy, getdist(xx, yy));
                        if(t.dis > res[i].dis) res[i] = t; 
                        else if(rand01() <= accept) res[i] = t;
                    }
                }
            }
            temper *= DEC;
            accept *= ACCEPT_DEC;
        }
        pa ans;
        ans.dis = 0;
        for (int i=1; i<=N; ++i)
            if(res[i].dis > ans.dis) ans = res[i];
        return ans;
    }
}

int main() {
    srand(19260817);
    int T; cin >> T;
    while(T--) {
        cin >> X >> Y >> n;
        for (int i=1; i<=n; ++i) scanf("%lf%lf", &a[i].x, &a[i].y);
        pa ans = SA::main();
        printf("The safest point is (%.1f, %.1f).\n", ans.x, ans.y);
    }
        
    return 0;
}

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