首页 > 代码库 > POJ 1379 Run Away 【基础模拟退火】
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;}
POJ 1379 Run Away 【基础模拟退火】
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