hdu1086 You can Solve a Geometry Problem too【排斥实验+跨立实验】,矩形,若两个矩形相交,则这两条线有可能相交跨立实验:通过叉乘判断两条线段是否相交,也就是一条线段的两个端点对另一条
Pointoperator- (constPoint&a) {returnPoint(x-a.x, y-a.y); } }; Point a1[maxn], a2[maxn];boolquick_exclude(Point a1, Point a2, Point b1, Point b2)//快速排斥实验{if(max(a1.x, a2.x) < min(b1.x, b2.x))returnfalse;if(max(b1.x, b2.x) < min(a1.x, a2.x))return...
hdu_1086 You can Solve a Geometry Problem too 题意:给定n条线段(n≤100),求有多少个交点(若有多于两条线段交于一点,应重复计数(即可以视为任意两条线段不交于同一点))。 题解:判定线段相交的裸题。只需要枚举任意两条线段是否相交即可。(方法为判断线段AB的端点A,B是否在线段CD两侧,且C,D在线段AB两侧...
HDU 1086 You can Solve a Geometry Problem too(判断线段相交),题目地址:HDU1086就这么一道仅仅判断线段相交的题目写了2k多B的代码。。是不是有点浪费。。。但是我觉得似乎哪里也优化是否在L1的两个端
向量判断HDU 1086 线段交点 You can Solve a Geometry Problem too 最近一直在学习向量判断之类的问题,现在正好有机会和大家讨论一下. 标题链接:http://acm.hdu.edu.cn/showproblem.php?pid=1086 分析:判断AB和CD两线段是否有交点: 同时满意两个条件,1.C点D点分别在AB的两......
//STATUS:C++_AC_15MS_204K #include<stdio.h> int is_inter(struct Segment a,struct Segment b); //跨立实验 int is_fxiji(struct Segment a,struct Segment b); //快速排斥 const int MAXN=100; struct Segment{ double x1,y1,x2,y2; }seg[MAXN]; int main() { int i,j,n,m; while(...
A. start over B. solve the problem C. make a guess D. do nothing 相关知识点: 试题来源: 解析 B。本题考查几何问题解决步骤。第一步画图表(draw a diagram),接着标注部分(label the parts),然后解决问题(solve the problem)。选项 A 重新开始错误。选项 C 瞎猜不对。选项 D 什么都不做也不对。
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Chapter 6.2 Notes: Use Proportions to Solve Geometry Problems Goal: You will use proportions to solve geometry problems. Additional Properties of Proportions: 1.Reciprocal Property: If two ratios are equal, then their reciprocals are also equal. 2.If you interchange the means of a proportion, the...
Solving one puzzle can give the compiler more clues on how to solve other puzzles Type inference errors If a puzzle can't be solved, there's an error in the source code During type inference, the compiler will record information about errors in source code Compiler uses heuristics to attemp...