HXX904的博客

题目描述An artist who wanted to create an installation where his works appeared to be floating in midair has cast a large cube of clear acrylic to serve as a base. Unfortunately, during the casting, some small flecks of dirt got into the mix, and now appear

问题：有一个玩具盒，被n个隔板分开成左到u右n+1个区域， 然后给每个玩具的坐标，求每个区域有几个玩具?补充一些数学知识：用向量叉积计算：设玩具坐标为(x,y),直线两端坐标为(x1,y1),(x2,y2); 设向量a坐标(x1-x,y1-y),向量b坐标(x2-x,y2-y)； a叉乘b等于((x1-x)*(y2-y))-((x2-x)*(y1-y)); 判断此叉积的正负即可。AC代码：/*有一个玩具盒，被n个隔板分开成左到u右n+1个区域，然后给每个玩具的坐标，求每个区域有.

原博客链接：https://www.cnblogs.com/tuyang1129/p/9390376.html。写的真是太好了，超赞！！！！问题：给出两条线段，问两线段是否相交？首先我们要明白一个定理：向量a×向量b（×为向量叉乘），若结果小于0，表示向量b在向量a的顺时针方向；若结果大于0，表示向量b在向量a的逆时针方向；若等于0，表示向量a与向量b平行。（顺逆时针是指两向量平移至起点相连，从某个方向旋转到另一个向量小于180度）。如下图：在上图中，OA×OB = 2 > 0，.

题意：给你一个10*10的正方形的盒子，里面有一些线段，让你求从坐标(0,5)出发到坐标(10,5)的最短路径。思路：判断任意两条线段的端点连线与其它线段是否有交点即可。最后用一个最短路径模板就可以了。补充两知识点：① 快速排斥实验：如果两条边所在矩形不相交则一定不相交，即满足设线段a为(x1,y1)--(x2,y2) 线段b为(x3,y3)--(x4,y4)这是判断可能相交的条件：min(x1,x2)<=max(x3,x4) && ...

问题：Once upon a time there was a greedy King who ordered his chief Architect to build a wall around the King's castle. The King was so greedy, that he would not listen to his Architect's proposals to...