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≤~

 

 1 /**************************************************************
 2     Problem: 3680
 3     User: rausen
 4     Language: C++
 5     Result: Accepted
 6     Time:300 ms
 7     Memory:1040 kb
 8 ****************************************************************/
 9  
10 #include <cstdio>
11 #include <cmath>
12  
13 using namespace std;
14 typedef double lf;
15 const int N = 10005;
16 const lf eps = 1e-5;
17  
18 #define P point
19 struct P {
20     lf x, y;
21     P() {}
22     P(lf _x, lf _y) : x(_x), y(_y) {}
23  
24     inline P operator * (lf X) {
25         return P(x * X, y * X);
26     }
27     inline P operator / (lf X) {
28         return P(x / X, y / X);
29     }
30     inline P operator + (P a) {
31         return P(x + a.x, y + a.y);
32     }
33      
34     inline bool operator == (P a) {
35         return fabs(x - a.x) < eps && fabs(y - a.y) < eps;
36     }
37     inline bool operator != (P a) {
38         return fabs(x - a.x) >= eps || fabs(y - a.y) >= eps;
39     }
40 }p[N];
41  
42 int n;
43 lf w[N];
44  
45 inline int read() {
46     int x = 0, sgn = 1;
47     char ch = getchar();
48     while (ch < '0' || '9' < ch) {
49         if (ch == '-') sgn = -1;
50         ch = getchar();
51     }
52     while ('0' <= ch && ch <= '9') {
53         x = x * 10 + ch - '0';
54         ch = getchar();
55     }
56     return sgn * x;
57 }
58  
59 inline lf sqr(lf x) {
60     return x * x;
61 }
62  
63 inline lf dist(P a, P b) {  
64     return sqrt(sqr(a.x - b.x) + sqr(a.y - b.y));
65 }
66  
67 int main() {
68     int i;
69     lf W = 0;
70     P p1 = P(0, 0), p2 = p1, p3;
71     n = read();
72     for (i = 1; i <= n; ++i) {
73         p[i].x = read(), p[i].y = read(), w[i] = read();
74         p1 = p1 + p[i] * w[i], W += w[i];
75     }
76     p1 = p1 / W;
77     while (p1 != p2) {
78         p2 = p3 = p1, p1 = P(0, 0);
79         W = 0;
80         for (i = 1; i <= n; ++i) {
81             p1 = p1 + p[i] * (w[i] / dist(p3, p[i]));
82             W += w[i] / dist(p3, p[i]);
83         }
84         p1 = p1 / W;    
85     }
86     printf("%.3lf %.3lf\n", p1.x, p1.y);
87     return 0;
88 }
View Code

（p.s. 这样迭代来做很快呢！）

posted on 2014-12-13 17:02  Xs酱~  阅读(313)  评论(0)  编辑  收藏  举报