[JSOI2004]平衡点 / 吊打XXX

题目链接

问题分析

随后系统的势能应当最低，即

\[\sum w_i \times \sqrt{(x-x_i)^2+(y-y_i)^2} \]

最小。直接模拟退火。

参考程序

#include <cstdio>
#include <ctime>
#include <cstdlib>
#include <cmath>

#define Maxn 1010
int n, x[Maxn], y[Maxn], w[Maxn];
double X, Y, Cost, AnsX, AnsY, AnsC;

inline double Rand() { return (double)rand() / RAND_MAX; }

inline double Calc(double X, double Y) {
	double Ans = 0;
	for (int i = 1; i <= n; ++i) 
		Ans += sqrt((x[i] - X) * (x[i] - X) + (y[i] - Y) * (y[i] - Y)) * w[i];
	if (Ans < AnsC) AnsX = X, AnsY = Y, AnsC = Ans;
	return Ans;
}

int main() {
	srand(time(NULL));
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i) scanf("%d%d%d", &x[i], &y[i], &w[i]);
	for (int i = 1; i <= n; ++i) AnsX += x[i], AnsY += y[i];
	AnsX = X = AnsX / n;
	AnsY = Y = AnsY / n;
	Cost = AnsC = Calc(X, Y);
	double T = 10000;
	while (T > 0.001) {
		double XX = X + T * (Rand() * 2 - 1);
		double YY = Y + T * (Rand() * 2 - 1);
		double C = Calc(XX, YY);
		if (exp((Cost - C) / T) > Rand()) X = XX, Y = YY, Cost = C;
		T *= 0.999;
	}
	for (int i = 1; i <= 1000; ++i) { //再来一点随机扰动
		double XX = AnsX + T * (Rand() * 2 - 1);
		double YY = AnsY + T * (Rand() * 2 - 1);
		Calc(XX, YY);
	}
	printf("%.3lf %.3lf\n", AnsX, AnsY);
	return 0;
}