[POJ][3285][Point of view in Flatland][模拟退火]

  
    

   
     /*
题目：Point of view in Flatland
题目来源：POJ 3285
题目难度：中等偏难
题目内容或思路：
 模拟退火
 这题的难点在找到合适的评估函数
做题日期：2011.3.9
*/
#include <cstdio>
#include <cstdlib>
#include <climits>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <queue>
#include <map>
#include <vector>
#include <bitset>
#include <cmath>
#include <set>
#include <utility>
#include <ctime>
#define sqr(x) ((x)*(x)) 
using namespace std;

const int N = 30, nt = 20, step = 30;
const double inf = 1e100;
const double eps = 1e-3;
const double pi = acos(-1.0);
double delta;

struct cpoint{
 double x, y, r;
}cp[N], test[N];

int dcmp(double x) {
 if (x < -eps) return -1; else return x > eps;
}

double dis(cpoint a, cpoint b) {
 return sqrt(sqr(a.x - b.x) + sqr(a.y - b.y));
}

void calc(cpoint &p) {
 double t[3], ave = 0;
 int id = 0;
 for (int i = 0; i < 3; ++i) {
 t[i] = dis(p, cp[i]) / cp[i].r;
 ave += t[i];
 }
 ave /= 3;
 p.r = 0;
 for (int i = 0; i < 3; ++i) {
 p.r += sqr(t[i] - ave);
 }
}

void updata(int id) {
 cpoint t;
 double angle = rand();
 angle -= int(angle / pi / 2) * pi * 2;
 t.x = test[id].x + delta * cos(angle);
 t.y = test[id].y + delta * sin(angle);
 calc(t);
 if (t.r < test[id].r)
 test[id] = t;
}

bool solve() {
 double t = 0, maxx, minx, maxy, miny;
 maxx = maxy = -inf;
 minx = miny = inf;
 for (int i = 0; i < 3; ++i) {
 scanf("%lf%lf%lf", &cp[i].x, &cp[i].y, &cp[i].r);
 t += cp[i].r;
 maxx = max(maxx, cp[i].x);
 minx = min(minx, cp[i].x);
 maxy = max(maxy, cp[i].y);
 miny = min(miny, cp[i].y);
 }
 if (dcmp(t) == 0) return false;
 for (int i = 0; i < nt; ++i) {
 test[i].x = minx + (rand() % (int)(maxx * 10 - minx * 10 + 1)) / 10.0;
 test[i].y = miny + (rand() % (int)(maxy * 10 - miny * 10 + 1)) / 10.0;
 calc(test[i]);
 }
 delta = max(maxx - minx, maxy - miny);
 for (; delta > 1e-6; delta *= 0.9) {
 for (int i = 0; i < nt; ++i)
 for (int j = 0; j < step; ++j)
 updata(i);
 }
 int res = -1;
 for (int i = 0; i < nt; ++i) {
 if (dcmp(test[i].r) == 0) {
 double u = dis(test[i], cp[0]);
 double v = dis(test[res], cp[0]);
 if (res == -1 || dcmp(u - v) < 0)
 res = i;
 }
 }
 if (res == -1) puts("No solution");
 else {
 if (dcmp(test[res].x) == 0) test[res].x = 0;
 if (dcmp(test[res].y) == 0) test[res].y = 0;
 printf("%.2lf %.2lf\n", test[res].x, test[res].y);
 }
 return true;
}

int main() {
#ifndef ONLINE_JUDGE
 freopen("D:\\in.txt", "r", stdin);
#endif
 while (solve());
 return 0;
}