USACO 5.2.2 fence3
题目：
http://ace.delos.com/usacoprob2?a=GPizy6nCKUy&S=fence3
http://pingce.ayyz.cn:9000/usaco/data/20110129214306/fence3.html
Kolstad & Schrijvers
Farmer John has decided to construct electric fences. He has fenced his fields into a number of bizarre shapes and now must find the optimal place to locate the electrical supply to each of the fences.
A single wire must run from some point on each and every fence to the source of electricity. Wires can run through other fences or across other wires. Wires can run at any angle. Wires can run from any point on a fence (i.e., the ends or anywhere in between) to the electrical supply.
Given the locations of all F (1 <= F <= 150) fences (fences are always parallel to a grid axis and run from one integer gridpoint to another, 0 <= X,Y <= 100), your program must calculate both the total length of wire required to connect every fence to the central source of electricity and also the optimal location for the electrical source.
The optimal location for the electrical source might be anywhere in Farmer John's field, not necessarily on a grid point.
PROGRAM NAME: fence3
INPUT FORMAT
The first line contains F, the number of fences.
F subsequent lines each contain two X,Y pairs each of which denotes the endpoints of a fence.
SAMPLE INPUT (file fence3.in)
3
0 0 0 1
2 0 2 1
0 3 2 3
OUTPUT FORMAT
On a single line, print three space-separated floating point numbers, each with a single decimal place. Presume that your computer's output library will round the number correctly.
The three numbers are:
- the X value of the optimal location for the electricity,
- the Y value for the optimal location for the electricity, and
- the total (minimum) length of the wire required.
SAMPLE OUTPUT (file fence3.out)
1.0 1.6 3.7
题解：
题意是在平面上找到一点使其到所给出的所有线段距离和最短，随机化算法即可，我写的模拟退火。注意由于最后对答案的精度要求不高，所以随机化算法的次数和步数都可以不太大，每次的步长也可以调大些。
View Code
1 /*
2 ID:zhongha1
3 PROB:fence3
4 LANG:C++
5 */
6
7 #include<cstdio>
8 #include<cstdlib>
9 #include<cstring>
10 #include<ctime>
11 #include<cmath>
12 #include<algorithm>
13
14 using namespace std;
15
16 const int TIME=1;
17 const double max_t=10000.0;
18 const double ratio=0.998;
19 const double per_t=25;
20 const double eps=1e-2;
21
22 int n;
23
24 double ans=1e+20,ansx,ansy,x11[201],x22[201],y11[201],y22[201];
25
26 double calc(double x,double y)
27 {
28 double v=0;
29 for (int a=1;a<=n;a++)
30 if (fabs(x11[a]-x22[a])<=eps)
31 {
32 double miny=min(y11[a],y22[a]);
33 double maxy=max(y11[a],y22[a]);
34 if (y<miny) v+=sqrt((x-x11[a])*(x-x22[a])+(y-miny)*(y-miny));
35 else
36 {
37 if (y>maxy) v+=sqrt((x-x11[a])*(x-x22[a])+(y-maxy)*(y-maxy));
38 else v+=fabs(x-x11[a]);
39 }
40 }
41 else
42 {
43 double minx=min(x11[a],x22[a]);
44 double maxx=max(x11[a],x22[a]);
45 if (x<minx) v+=sqrt((y-y11[a])*(y-y22[a])+(x-minx)*(x-minx));
46 else
47 {
48 if (x>maxx) v+=sqrt((y-y11[a])*(y-y22[a])+(x-maxx)*(x-maxx));
49 else v+=fabs(y-y11[a]);
50 }
51 }
52 return v;
53 }
54
55 void solve()
56 {
57 for (int a=1;a<=TIME;a++)
58 {
59 double nowx=rand()%110;
60 double nowy=rand()%110;
61 double nowv=calc(nowx,nowy);
62 for (double t=max_t;t>eps;t*=ratio)
63 {
64 for (int b=1;b<=per_t;b++)
65 {
66 double deltax=(double)(rand()%100)/100.0;
67 double deltay=(double)(rand()%100)/100.0;
68 if (rand()%2==0) deltax=-deltax;
69 if (rand()%2==0) deltay=-deltay;
70 double newv=calc(nowx+deltax,nowy+deltay);
71 double delta=nowv-newv;
72 if (newv<nowv || (double)rand()/RAND_MAX<exp((double)delta))
73 {
74 nowv=newv;
75 nowx=nowx+deltax;
76 nowy=nowy+deltay;
77 }
78 if (nowv<ans)
79 {
80 ans=nowv;
81 ansx=nowx;
82 ansy=nowy;
83 }
84 }
85 }
86 }
87 }
88
89 int main()
90 {
91 freopen("fence3.in","r",stdin);
92 freopen("fence3.out","w",stdout);
93
94 srand(time(0));
95 scanf("%d",&n);
96 for (int a=1;a<=n;a++)
97 scanf("%lf%lf%lf%lf",&x11[a],&y11[a],&x22[a],&y22[a]);
98 solve();
99 printf("%.1lf %.1lf %.1lf\n",ansx,ansy,ans);
100
101 return 0;
102 }
