水题直接码... /********************* Template ************************/ #include <set> #include <map> #include <list> #include <cmath> #include <ctime> #include <deque> #include <queue> #include <stack> #include…

/* poj 1654 Area 多边形面积 题目意思很简单,但是1000000的point开不了 */ #include<stdio.h> #include<math.h> #include<string.h> const int N=1000000+10; const double eps=1e-8; struct point { double x,y; point(){} point(double a,double b):x(a),y(b){} }; int le…

#include<stdio.h> #include<algorithm> #include <cstring> using namespace std; typedef long long ll; const int MAXN = 1000008; char s[MAXN]; int dx[] = {-1, -1, -1, 0, 0, 0, 1, 1, 1}; int dy[] = {-1, 0, 1, -1, 0, 1, -1, 0, 1}; int main()…

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…

链接:http://poj.org/problem?id=1654 Area Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 14952   Accepted: 4189 Description You are going to compute the area of a special kind of polygon. One vertex of the polygon is the origin of the orth…

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 +…

Area Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 17456   Accepted: 4847 Description You are going to compute the area of a special kind of polygon. One vertex of the polygon is the origin of the orthogonal coordinate system. From thi…

Area Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 16894   Accepted: 4698 Description You are going to compute the area of a special kind of polygon. One vertex of the polygon is the origin of the orthogonal coordinate system. From thi…

一个简单的用叉积求任意多边形面积的题,并不难,但我却错了很多次,double的数据应该是要转化为long long,我转成了int...这里为了节省内存尽量不开数组,直接计算,我MLE了一发...,最后看了下别人的才过,我的代码就不发了,免得误导,不得不说几何真是... 还有就是这个大神的代码,貌似G++,过不了,C++AC #include <iostream> #include <algorithm> #include <cstdio> #include <c…

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…