Scrambled Polygon Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 7799   Accepted: 3707 Description A closed polygon is a figure bounded by a finite number of line segments. The intersections of the bounding line segments are called the…

/** 极角排序输出,,, 主要atan2(y,x) 容易失精度,,用 bool cmp(point a,point b){ 5 if(cross(a-tmp,b-tmp)>0) 6 return 1; 7 if(cross(a-tmp,b-tmp)==0) 8 return length(a-tmp)<length(b-tmp); 9 return 0; 10 } **/ #include <iostream> #include <algorithm> #includ…

Scrambled Polygon Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 7805   Accepted: 3712 Description A closed polygon is a figure bounded by a finite number of line segments. The intersections of the bounding line segments are called the…

Scrambled Polygon Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 8636   Accepted: 4105 Description A closed polygon is a figure bounded by a finite number of line segments. The intersections of the bounding line segments are called the…

Scrambled Polygon Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 7214   Accepted: 3445 Description A closed polygon is a figure bounded by a finite number of line segments. The intersections of the bounding line segments are called the…

1007: [HNOI2008]水平可见直线 Time Limit: 1 Sec  Memory Limit: 162 MBSubmit: 2605  Solved: 914[Submit][Status] Description Input 第一行为N(0 < N < 50000),接下来的N行输入Ai,Bi Output 从小到大输出可见直线的编号,两两中间用空格隔开,最后一个数字后面也必须有个空格 Sample Input 3-1 01 00 0 Sample Output 1 2 HI…

http://poj.org/problem?id=2007 Time Limit: 1000MS   Memory Limit: 30000K Total Submissions: 6701   Accepted: 3185 Description A closed polygon is a figure bounded by a finite number of line segments. The intersections of the bounding line segments ar…

题目传送门 题意:裸的对原点的极角排序,凸包貌似不行. /************************************************ * Author :Running_Time * Created Time :2015/11/3 星期二 14:46:47 * File Name :POJ_2007.cpp ************************************************/ #include <cstdio> #include <al…

题目链接 题意 : 对输入的点极角排序 思路 : 极角排序方法 #include <iostream> #include <cmath> #include <stdio.h> #include <algorithm> using namespace std; struct point { double x,y; }p[],pp; double cross(point a,point b,point c) { return (a.x-c.x)*(b.y-c.y…

水题,根本不用凸包,就是一简单的极角排序. 叉乘<0,逆时针. #include <iostream> #include <cstdio> #include <cstring> #include <string> #include <algorithm> using namespace std; const int maxn=55; struct point { double x,y; } p[maxn]; double cross(poi…