求凸包面积.求结果后不用加绝对值,这是BBS()排序决定的. //Ps 熟练了template <class T>之后用起来真心方便= = //POJ 3348 //凸包面积 //1A 2016-10-15 #include <cstdio> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #define MAXN (10000 +…

Cows Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 8122   Accepted: 3674 Description Your friend to the south is interested in building fences and turning plowshares into swords. In order to help with his overseas adventure, they are f…

Cows Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 6199   Accepted: 2822 Description Your friend to the south is interested in building fences and turning plowshares into swords. In order to help with his overseas adventure, they are f…

自己想了一个方法判断点是不是在凸包内,先求出凸包面积,在求由点与凸包上每两个点之间的面积(点已经排好序了),如果两者相等,则点在凸包内,否则不在(时间复杂度可能有点高)但是这题能过 #include<map> #include<set> #include<cmath> #include<queue> #include<stack> #include<vector> #include<cstdio> #include<…

用叉积求凸包面积 如图所示,每次找p[0]来计算,(叉积是以两个向量构成的平行四边形的面积,所以要/2) #include<map> #include<set> #include<cmath> #include<queue> #include<stack> #include<vector> #include<cstdio> #include<cassert> #include<iomanip> #i…

Cows Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 8063   Accepted: 3651 Description Your friend to the south is interested in building fences and turning plowshares into swords. In order to help with his overseas adventure, they are f…

点此看题面 大致题意: 告诉你若干个矩形的重心坐标.长.宽和相对\(y\)轴的偏转角度,求矩形面积和与能围住这些矩形的最小凸包面积之比. 矩形面积和 这应该是比较好求的吧. 已经给了你长和宽,直接乘起来累加即可. 最小凸包面积 这道题关键还是在于求凸包面积. 首先,我们要注意将题目中给出的角度转换成弧度,还要记得取相反数,不然调死你. 这段代码可以与旋转函数放在一起: inline Point Rotate(Vector A,double deg) { register double rad=d…

题目链接:https://vjudge.net/problem/POJ-3348 题意:转换题意后即是求凸包的面积. 思路: 套模板,求凸包面积即转换为多个三角形面积之和,用叉积求,然后除2,因为本题除50,所以最后除100. AC code: #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> using namespace std; ; const double P…

type node=record x,y:longint; end; ; var k,q,qq:longint; sum:double; f,g:..maxn] of node; m,i,j,a,b:longint; stack:..maxn] of longint; nm:longint; function dis(a,b:node):double; begin exit(sqrt(sqr(a.x-b.x)+sqr(a.y-b.y))); end; procedure swap(var a,b…

题意: 求一条直线分凸包两边的面积. 解法: 因为题意会说一定穿过,那么不会有直线与某条边重合的情况.我们只要找到一个直线分成的凸包即可,另一个的面积等于总面积减去那个的面积. 怎么得到分成的一个凸包呢? 从0~n扫过去,如果扫到的边与直线不相交,那么把端点加进新凸包中,如果直线与扫到的边相交了,那么就将交点加入新凸包,然后以后不相交的话也不加入点到新凸包中,直到遇到下一个与直线相交的边,则把交点又加入新凸包,然后在扫到末尾加入点.这样就得到了. 即找到如图: 注意四舍五入. 代码: #incl…