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我也来说几句我的想法注意输入,前面有人说了 1.排除线段在矩形上下左右四周的情况,F 2.排除线段在矩形内部的情况,T 3.调用两次检查两个点是否分在直线两侧,即将两个点的坐标代入直线方程 bool f(int a,int b,int c,int d) //(a,b) (c,d)两点 { int p=(y2-y1)*(a-x1)-(b-y1)*(x2-x1); int q=(y2-y1)*(c-x1)-(d-y1)*(x2-x1); if(p*q>0) return false; //相交 else return true; //不相交 } Followed by: Post your reply here: |
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