BZOJ 3680: 吊打XXX （模拟退火）

//yy：今天简单入门学了下ORZ

爬山算法：兔子朝着比现在高的地方跳去。它找到了不远处的最高山峰。但是这座山不一定是珠穆朗玛峰。这就是爬山算法，它不能保证局部最优值就是全局最优值。

模拟退火：兔子喝醉了。它随机地跳了很长时间。这期间，它可能走向高处，也可能踏入平地。但是，它渐渐清醒了并朝最高方向跳去。这就是模拟退火。

题目链接：BZOJ 3680: 吊打XXX

1<=n<=10000,-100000<=xi,yi<=100000

题意：找一个点，使得最小

 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <cmath>
 #include <climits>
 #include <cstdlib>
 #define FIRE(x) (x *= 0.98)
 using namespace std;
 typedef long long ll;
 const int N = ;
 const double inf = 1e17;
 const double PI = acos(-1.0);
 const double eps = 1e-;
 struct Point{
     double x, y, w;
     Point(){}
     Point(double _x, double _y):x(_x), y(_y) {}
     Point operator -(const Point &b)const{
         return Point(x - b.x,y - b.y);
     }
     double operator *(const Point &b)const{
         return x*b.x + y*b.y;
     }
 }now, ans, po[N];
 double dist(Point a, Point b) {return sqrt((a-b)*(a-b));}
 int n;
 double tot = inf;
 double statistic(Point p) {
     double res = 0.0;
     for(int i = ; i < n; ++i) res += dist(p, po[i]) * po[i].w;
     if(res < tot) tot = res, ans = p;
     return res;
 }
 double Rand() {return (rand()% + ) / 1000.0;}
 void SA(double T) {
     double alpha, sub;
     while(T > eps) {
         alpha = 2.0 * PI * Rand();
         Point t(now.x + T * cos(alpha), now.y + T * sin(alpha));
         sub = statistic(now) - statistic(t);
         if(sub >=  || exp(sub / T) >= Rand()) now = t;
         FIRE(T);
     }
     T = 0.001;
     for(int i = ; i <= ; ++i) {
         alpha = 2.0 * PI * Rand();
         Point t(ans.x + T * cos(alpha) * Rand(), ans.y + T * sin(alpha) * Rand());
         statistic(t);
     }
 }
 int main(){
     srand();
     scanf("%d", &n);
     for(int i = ; i < n; ++i) {
         scanf("%lf%lf%lf", &po[i].x, &po[i].y, &po[i].w);
         now.x += po[i].x;   now.y += po[i].y;
     }
     now.x /= n; now.y /= n;
     double T = 1000.0;
     SA(T);
     printf("%.3f %.3f\n", ans.x, ans.y);
     return ;
 }