• poj1379 Run Away


    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;
        }
        inline pa randpoint(double px, double py, double qx, double qy) {
            double x = (qx - px) * rand01() + px, y = (qy - py) * rand01() + py;
            return pa(x, y, getdist(x, y)); 
        }
        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) {
                        pa t = randpoint(max(res[i].x - temper, 0.0), max(res[i].y - temper, 0.0), min(res[i].x + temper, X), min(res[i].y + temper, Y));
                        if(0 <= t.x && t.x <= X && 0 <= t.y && t.y <= Y) {
                            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).
    ", ans.x, ans.y);
        }
            
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/galaxies/p/poj1379.html
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