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; }
时间: 2024-12-19 01:36:33