BZOJ3680 吊打XXX

大家都是用什么爬山算法、模拟退火算法的。。。太高端了蒟蒻不会

于是Xs找到了些奇怪的东西

这篇论文的3.3节 "N孔系统"就是这道题呢~

于是就没啦≥v≤~

/**************************************************************
    Problem: 3680
    User: rausen
    Language: C++
    Result: Accepted
    Time:300 ms
    Memory:1040 kb
****************************************************************/
 
#include <cstdio>
#include <cmath>
 
using namespace std;
typedef double lf;
const int N = 10005;
const lf eps = 1e-5;
 
#define P point
struct P {
    lf x, y;
    P() {}
    P(lf _x, lf _y) : x(_x), y(_y) {}
 
    inline P operator * (lf X) {
        return P(x * X, y * X);
    }
    inline P operator / (lf X) {
        return P(x / X, y / X);
    }
    inline P operator + (P a) {
        return P(x + a.x, y + a.y);
    }
     
    inline bool operator == (P a) {
        return fabs(x - a.x) < eps && fabs(y - a.y) < eps;
    }
    inline bool operator != (P a) {
        return fabs(x - a.x) >= eps || fabs(y - a.y) >= eps;
    }
}p[N];
 
int n;
lf w[N];
 
inline int read() {
    int x = 0, sgn = 1;
    char ch = getchar();
    while (ch < '0' || '9' < ch) {
        if (ch == '-') sgn = -1;
        ch = getchar();
    }
    while ('0' <= ch && ch <= '9') {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return sgn * x;
}
 
inline lf sqr(lf x) {
    return x * x;
}
 
inline lf dist(P a, P b) {  
    return sqrt(sqr(a.x - b.x) + sqr(a.y - b.y));
}
 
int main() {
    int i;
    lf W = 0;
    P p1 = P(0, 0), p2 = p1, p3;
    n = read();
    for (i = 1; i <= n; ++i) {
        p[i].x = read(), p[i].y = read(), w[i] = read();
        p1 = p1 + p[i] * w[i], W += w[i];
    }
    p1 = p1 / W;
    while (p1 != p2) {
        p2 = p3 = p1, p1 = P(0, 0);
        W = 0;
        for (i = 1; i <= n; ++i) {
            p1 = p1 + p[i] * (w[i] / dist(p3, p[i]));
            W += w[i] / dist(p3, p[i]);
        }
        p1 = p1 / W;    
    }
    printf("%.3lf %.3lf
", p1.x, p1.y);
    return 0;
}

（p.s. 这样迭代来做很快呢！）