题目:平面上给定n条线段,找出一个点,使这个点到这n条线段的距离和最小。
源码如下:
#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <stdio.h>
#include <time.h>
#include <math.h> #define N 1005
#define eps 1e-8 //搜索停止条件阀值
#define INF 1e99
#define delta 0.98 //温度下降速度
#define T 100 //初始温度 using namespace std; int dx[] = {, , -, };
int dy[] = {-, , , }; //上下左右四个方向 struct Point
{
double x, y;
}; Point s[N], t[N]; double cross(Point A, Point B, Point C)
{
return (B.x - A.x) * (C.y - A.y) - (B.y - A.y) * (C.x - A.x);
} double multi(Point A, Point B, Point C)
{
return (B.x - A.x) * (C.x - A.x) + (B.y - A.y) * (C.y - A.y);
} double dist(Point A, Point B)
{
return sqrt((A.x - B.x) * (A.x - B.x) + (A.y - B.y) * (A.y - B.y));
} double GetDist(Point A, Point B, Point C)
{
if(dist(A, B) < eps) return dist(B, C);
if(multi(A, B, C) < -eps) return dist(A, C);
if(multi(B, A, C) < -eps) return dist(B, C);
return fabs(cross(A, B, C) / dist(A, B));
} double GetSum(Point s[], Point t[], int n, Point o)
{
double ans = ;
while(n--)
ans += GetDist(s[n], t[n], o);
return ans;
} double Search(Point s[], Point t[], int n, Point &o)
{
o = s[];
double tem = T;
double ans = INF;
while(tem > eps)
{
bool flag = ;
while(flag)
{
flag = ;
for(int i = ; i < ; i++) //上下左右四个方向
{
Point z;
z.x = o.x + dx[i] * tem;
z.y = o.y + dy[i] * tem;
double tp = GetSum(s, t, n, z);
if(ans > tp)
{
ans = tp;
o = z;
flag = ;
}
}
}
tem *= delta;
}
return ans;
} int main()
{
int n;
while(scanf("%d", &n) != EOF)
{
for(int i = ; i < n; i++)
scanf("%lf %lf %lf %lf", &s[i].x, &s[i].y, &t[i].x, &t[i].y);
Point o;
double ans = Search(s, t, n, o);
printf("%.1lf %.1lf %.1lf\n", o.x, o.y, ans);
}
return ;
}