UVa 10263

题目：已知一个点M，以及一个由N条线段构成的曲线（N+1个点），求曲线上到M最近的点坐标。

分析：计算几何。点到线段距离。

            点到线短距离：1.求点到直线距离；

                                        2.判断垂足是否在线段上：在则为答案；不在则端点为答案；

            枚举线段，找到最小距离对应的点即可；

说明：雅思考了6.0，还点再考~~~~(>_<)~~~~。

#include <stdio.h>#include <stdlib.h>#include <math.h>const double eps = 1e-6;typedef struct _point{double x, y;}point;point M, P1, P2;point new_point(double x, double y){point ans;ans.x = x;ans.y = y;return ans;}double distance(point p, point q){return sqrt((p.x-q.x)*(p.x-q.x) + (p.y-q.y)*(p.y-q.y));}point foot_point(point m, point p, point q) // Ax + by + C = 0{double A1 = p.x-q.x, B1 = p.y-q.y, C1 = (q.y-p.y)*m.y+(q.x-p.x)*m.x;double A2 = q.y-p.y, B2 = p.x-q.x, C2 = (q.x-p.x)*p.y+(p.y-q.y)*p.x;double x = (C2*B1-C1*B2)/(A1*B2-A2*B1);     double y = (C2*A1-C1*A2)/(A2*B1-A1*B2);return new_point(x, y);}bool on_line(point m, point p, point q){return (p.x-m.x)*(q.x-m.x) <= eps && (p.y-m.y)*(q.y-m.y) <= eps;}point closest_point_on_segement(point m, point p, point q){point ans = foot_point(m, p, q);if (!on_line(ans, p, q)) {if (distance(m, p) - distance(m, q) > eps) {return q;}else {return p;}}else {return ans;}}int main(){int n;while (~scanf("%lf%lf", &M.x, &M.y)) {scanf("%d%lf%lf", &n, &P1.x, &P1.y);point  ans = P1;double min_dist = distance(M, P1);while (n --) {scanf("%lf%lf", &P2.x, &P2.y);point Q = closest_point_on_segement(M, P1, P2);if (distance(M, ans) - distance(M, Q) > eps) {ans = Q;}P1 = P2;}printf("%.4lf\n%.4lf\n", ans.x, ans.y);}return 0;}