[JSOI2004]平衡点 / 吊打XXX

Posted 2021-03-23 chy-2003

题目链接

问题分析

随后系统的势能应当最低，即

[ sum w_i imes 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
", AnsX, AnsY);
    return 0;
}