carter91620

//分别计算每个圆在floor上的阴影部分，然后贪心一下就好了//在求切线时，是把圆点逆时针旋转angle和2*pi-angle度#include#include#include#include#include#define dist(a,b) sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y))#define cross(a,b,c)

//模拟退火算法//1、随机选取一个足够大的温度T作为开始//2、随机选取P个起始点，作为可行解//3、分别更新这P个可行解//4、减小温度T，直到终止条件//5、转到1#include#include#include#include#include#define dist(a,b) sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y))

//半平面交，然后求凸包对重点#include#include#include#include#define eps 1e-8#define dist(a,b) sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y))#define cross(a,b,c) (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x)#defi

//简单题，按题目要求求就行。#include#include#include#include#define eps 1e-8#define dist(a,b) sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y))#define cross(a,b,c) (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x)#defi

//对n*n个四边形，求面积最大的那个的面积#include#include#include#include#define eps 1e-8#define dist(a,b) sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y))#define cross(a,b,c) (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x

//枚举第i和第j个转折点，求最大距离#include#include#include#include#define eps 1e-8#define dist(a,b) sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y))#define cross(a,b,c) (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x)#

//由题目给出的公式，积分//其实我也是看网上的结论这题#include#include#include#include#include#define dist(a,b) sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y))#define cross(a,b,c) (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x

//求最少的边长把城池围住//其实就是求凸包的周长+圆周长#include#include#include#include#include#define dist(a,b) sqrt(1.0*(a.x-b.x)*(a.x-b.x)+1.0*(a.y-b.y)*(a.y-b.y))#define cross(a,b,c) 1.0*(b.x-a.x)*(c.y-a.y)-1.0

//半平面交，就是求有没有内核，有输出“YES”，否则“NO”//这里都是逆时针方向#include#include#include#include#include#define dist(a,b) sqrt(1.0*(a.x-b.x)*(a.x-b.x)+1.0*(a.y-b.y)*(a.y-b.y))#define cross(a,b,c) 1.0*(b.x-a.x)*(c

//运用叉积判断是否是逆时针，因为同一直线上也行，所以有n个植物就可以吃掉n个植物#include#include#include#include#include#define dist(a,b) sqrt(1.0*(a.x-b.x)*(a.x-b.x)+1.0*(a.y-b.y)*(a.y-b.y))#define cross(a,b,c) 1.0*(b.x-a.x)*(c.y-

//无比之水，按照题目要求做就行，求面积时，把面积之和后，在除2，避免精度缺失。#include#include#include#include#include#define dist(a,b) sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y))#define cross(a,b,c) (b.x-a.x)*(c.y-a.y)-(b.y-a.y

//这道题就是麻烦了点，其他都很简单，细节问题。#include#include#include#includeusing namespace std;#define PR 1e-8#define N 10005double edge[N][N];double dis[N];bool vis[N];struct TPoint{ double x,y; TPoint

//从1到n枚举，求砍到最少的树//位压缩，之开始用的IDA*一直WA，遂换成位压缩果断AC#include#include#include#include#define eps 1e-8#define dist(a,b) sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y))#define cross(a,b,c) (b.x-a.x)*(c.y

//把相交的矩形分成若干个小矩形//vis[i][j]表示由标号为xi和yj的小矩形是不是原来矩形的一部分#include#include#include#include#include#define dist(a,b) sqrt(1.0*(a.x-b.x)*(a.x-b.x)+1.0*(a.y-b.y)*(a.y-b.y))#define cross(a,b,c) 1.0*(

//求在可以旋转的给定圆心和半径的半圆中最多点的个数#include#include#include#include#include#define dist(a,b) sqrt(1.0*(a.x-b.x)*(a.x-b.x)+1.0*(a.y-b.y)*(a.y-b.y))#define cross(a,b,c) 1.0*(b.x-a.x)*(c.y-a.y)-1.0*(b.y-a

公式公式，公式的水题。。。。#include#include#include#includeusing namespace std;#define pi acos(-1.0)#define eps 1e-8struct TPoint{ double al,bk;}p1,p2,p3;double getangle(TPoint a,TPoint b){ double a