HDU3694 Fermat Point in Quadrangle

　　原题传送：http://acm.hdu.edu.cn/showproblem.php?pid=3694

　　题意很简单，求平面四个点的费马点。

这道题直接枚举所有情况：

　　1.四个点独立不重叠，这种情况又分为两种情况：

　　　　a.四个点构成凸四边形，那么对角线交点就是费马点

　　　　b.四个点构成凹四边形，那么费马点肯定是凹的那个点 （因为漏了这个情况wa了10多次）

　　2.四个点中有重复的点，费马点就是重复的那个点

View Code

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <stdlib.h>
#define INF 10000000.0
const double eps = 1e-6;

struct point{double x, y;};

inline double min(double a, double b){return a < b ? a : b;}

inline double max(double a, double b){return a > b ? a : b;}

bool inter(point a, point b, point c, point d)  // 判相交
{
if(min(a.x, b.x) > max(c.x, d.x) ||
min(a.y, b.y) > max(c.y, d.y) ||
min(c.x, d.x) > max(a.x, b.x) ||
min(c.y, d.y) > max(a.y, b.y) )
return 0;

double h, i, j, k;
h = (b.x - a.x) * (c.y - a.y) - (b.y - a.y) * (c.x - a.x);
i = (b.x - a.x) * (d.y - a.y) - (b.y - a.y) * (d.x - a.x);
j = (d.x - c.x) * (a.y - c.y) - (d.y - c.y) * (a.x - c.x);
k = (d.x - c.x) * (b.y - c.y) - (d.y - c.y) * (b.x - c.x);
return h * i <= eps && j * k <= eps;
}

double cal(point a, point b)      // 计算两点距离
{
return sqrt((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y));
}

int main()
{
point a, b, c, d;
double len[20];
while(scanf("%lf%lf%lf%lf%lf%lf%lf%lf", &a.x, &a.y, &b.x, &b.y, &c.x, &c.y, &d.x, &d.y))
{
if(a.x == -1)
break;
for(int i = 1; i < 20; i ++)
len[i] = INF;

if(a.x == d.x && a.y == d.y)
len[1] = cal(a, b) + cal(a, c); //a,d重合
else if(a.x == c.x && a.y == c.y)
len[2] = cal(a, b) + cal(a, d); //a,c重合
else if(a.x == b.x && a.y == b.y)
len[3] = cal(a, c) + cal(a, d); //a,b重合
else if(b.x == c.x && b.y == c.y)
len[4] = cal(b, a) + cal(b, d); //b,c重合
else if(b.x == d.x && b.y == d.y)
len[5] = cal(b, a) + cal(b, c); //b,d重合
else if(c.x == b.x && c.y == b.y)
len[6] = cal(c, a) + cal(c, b); //c,d重合
else
{
if(inter(a, b, c, d))           //ab, cd相交
len[7] = cal(a, b) + cal(c, d);
else if(inter(a, c, b, d))      //ac, bd相交
len[8] = cal(a, c) + cal(b, d);
else if(inter(a, d, b, c))      //ad, bc相交
len[9] = cal(a, d) + cal(b, c);
}
len[10] = cal(a, b) + cal(a, c) + cal(a, d); // 处理凹多边形
len[11] = cal(b, a) + cal(b, c) + cal(b, d);
len[12] = cal(c, a) + cal(c, b) + cal(c, d);
len[13] = cal(d, a) + cal(d, b) + cal(d, c);
double m = INF;
for(int i = 1; i < 20; i ++)
m = min(len[i], m);
printf("%.4lf\n", m);
}
return 0;
}