摘要: 这个题就是纯暴力枚举:枚举所有顶点的连线的直线与所有线段是否有交点;View Code #include<iostream>#include<cstdio>#include<cstdlib>#include<algorithm>#include<cmath>#include<queue>#include<set>#include<map>#include<cstring>#include<vector>using namespace std;class Line{public 阅读全文
posted @ 2012-07-11 22:24 wutaoKeen 阅读(97) 评论(0) 推荐(0) 编辑
摘要: 这个题就是求凸包;极角排序:View Code View Code #include<iostream>#include<cstdio>#include<cstdlib>#include<algorithm>#include<cmath>#include<queue>#include<set>#include<map>#include<cstring>#include<vector>using namespace std;class Point{public: double 阅读全文
posted @ 2012-07-11 21:19 wutaoKeen 阅读(120) 评论(0) 推荐(0) 编辑
摘要: 这个题就是求无论你站在家的那个地方都能看到财产的最大值;这个题可以从正面做,直接用叉积求出最大值;View Code #include<iostream>#include<cstdio>#include<cstdlib>#include<algorithm>#include<cmath>#include<cstring>using namespace std;class Line{public: double x1,x2,y; }line[10000];Line house,pro;int dcmp( double x ) 阅读全文
posted @ 2012-07-11 17:52 wutaoKeen 阅读(165) 评论(0) 推荐(0) 编辑
摘要: 这个题首先判断是否是凸多边形,再判断圆心是否在凸多边形内,再判断凸多边形是否在圆内;做法用Graham判断是否是多边形;再把圆心加入到多边形的点之中,再用Graham看圆心是否在凸多边形上,如果在就代表圆心在凸多边形外;再用点到直线的距离,fabs(A*x + B*y + c)/sqrt(A*A + B*B);极角排序法:View Code #include<iostream>#include<cstdio>#include<cstdlib>#include<algorithm>#include<cmath>#include<q 阅读全文
posted @ 2012-07-11 09:14 wutaoKeen 阅读(190) 评论(0) 推荐(0) 编辑