地址：https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=3799Two star-crossed lovers want to meet. The two lovers are standing at distinct points in the plane (but then again, aren‘t we all?). They can travel freely except that there is a single wall which cannot be crossed. The wall is a line segment which is parallel to either the x or y axis. Each lover can move 1 unit in 1 second. How long will it take them to be together if they both choose the best path?
输入
Input for each test case will consist of two lines each containing four integers. The first two integers will specify the x and y coordinates of the first lover; the next two integers will specify the x and ycoordinates of the second lover. The next four integers will specify the start and end points of the wall. Furthermore, in all cases both lovers will not be on the (infinite) line containing the wall -- that is, the wall extended in both directions. All coordinates will be positive and less than or equal to 10000 and neither lover will start on the wall. The input will be terminated by a line containing four zeroes.
输出
For each test case, output the minimum time in seconds for the two lovers to meet. Print the answer to exactly 3 decimal places, using the output format shown in the example.、
输入
5 2 7 2 1 1 1 100 1 2 3 2 2 1 2 100 0 0 0 0输出
Case 1: 1.000 Case 2: 1.414题意： A为起点，B为终点。要求从A走到B不能直接通过线段CD的最短路程。解题思路：1、AB与CD不相交。ANS= AB2、AB与CD相交。ANS=MIN(AC+BC,AD+BD)代码：#include <stdio.h> #include <math.h> struct point { double x; double y; }; double direction(point p1, point p2, point p) { return (p1.x - p.x)*(p2.y - p.y) - (p2.x - p.x)*(p1.y - p.y); } int on_segment(point p1, point p2, point p) { double max = p1.x > p2.x ? p1.x : p2.x; double min = p1.x < p2.x ? p1.x : p2.x; if (p.x >= min && p.x <= max) return 1; else return 0; } // 判断线段p1p2是否与p3p4相交 int segments_intersert(point p1, point p2, point p3, point p4) { double d1, d2, d3, d4; d1 = direction(p1, p2, p3); d2 = direction(p1, p2, p4); d3 = direction(p3, p4, p1); d4 = direction(p3, p4, p2); if (d1*d2 < 0 && d3*d4 < 0) return 1; else if (d1 == 0 && on_segment(p1, p2, p3)) return 1; else if (d2 == 0 && on_segment(p1, p2, p4)) return 1; else if (d3 == 0 && on_segment(p3, p4, p1)) return 1; else if (d4 == 0 && on_segment(p3, p4, p2)) return 1; return 0; } int main() { point a, b, c, d; int C = 1; while (scanf("%lf%lf%lf%lf", &a.x, &a.y, &b.x, &b.y) != EOF) { double ans; if (a.x == 0 && a.y == 0 && b.x == 0 && b.y == 0) return 0; scanf("%lf%lf%lf%lf", &c.x, &c.y, &d.x, &d.y); if (segments_intersert(a, b, c, d)){ double ac = sqrt((a.x - c.x)*(a.x - c.x) + (a.y - c.y)*(a.y - c.y)); double bc = sqrt((b.x - c.x)*(b.x - c.x) + (b.y - c.y)*(b.y - c.y)); double ad = sqrt((a.x - d.x)*(a.x - d.x) + (a.y - d.y)*(a.y - d.y)); double bd = sqrt((b.x - d.x)*(b.x - d.x) + (b.y - d.y)*(b.y - d.y)); ans = ac + bc; if (ans > (ad + bd)) ans = ad + bd; } else ans = sqrt((a.x - b.x)*(a.x - b.x) + (a.y - b.y)*(a.y - b.y)); printf("Case %d: %.3lf\n", C++, ans / 2.0); } return 0; }
原文地址：http://blog.csdn.net/u012964281/article/details/38056951
