FZU 2148 Moon Game
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Description
Fat brother and Maze are playing a kind of special (hentai) game in the clearly blue sky which we can just consider as a kind of two-dimensional plane. Then Fat brother starts to draw N starts in the sky which we can just consider each as a point. After he draws these stars, he starts to sing the famous song “The Moon Represents My Heart” to Maze.
You ask me how deeply I love you,
How much I love you?
My heart is true,
My love is true,
The moon represents my heart.
…
But as Fat brother is a little bit stay-adorable(呆萌), he just consider that the moon is a special kind of convex quadrilateral and starts to count the number of different convex quadrilateral in the sky. As this number is quiet large, he asks for your help.
Input
The first line of the date is an integer T, which is the number of the text cases.
Then T cases follow, each case contains an integer N describe the number of the points.
Then N lines follow, Each line contains two integers describe the coordinate of the point, you can assume that no two points lie in a same coordinate and no three points lie in a same line. The coordinate of the point is in the range[-10086,10086].
1 <= T <=100, 1 <= N <= 30
Output
For each case, output the case number first, and then output the number of different convex quadrilateral in the sky. Two convex quadrilaterals are considered different if they lie in the different position in the sky.
Sample Input
Sample Output
#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
using namespace std; int T;
int n,ca=;
int i,j,k,l;
double x[],y[]; double S(double x1,double y1,double x2,double y2,double x3,double y3)
{
double K=(x1*y2+x2*y3+x3*y1-x1*y3-x2*y1-x3*y2)/2.0;
return fabs(K);
} int check()
{
double S123,S124,S134,S234;
S123=S(x[i],y[i],x[j],y[j],x[k],y[k]);
S124=S(x[i],y[i],x[j],y[j],x[l],y[l]);
S134=S(x[i],y[i],x[k],y[k],x[l],y[l]);
S234=S(x[j],y[j],x[k],y[k],x[l],y[l]);
//printf("%lf %lf %lf %lf\n",S123,S124,S134,S234);
if(fabs(S124+S134+S234-S123)==)
return ;
if(fabs(S124+S134-S234+S123)==)
return ;
if(fabs(S124-S134+S234+S123)==)
return ;
if(fabs(S134+S234+S123-S124)==)
return ;
return ;
} int main()
{
scanf("%d",&T);
while(T--)
{
int num=;
scanf("%d",&n);
for(i=;i<=n;i++)
scanf("%lf %lf",&x[i],&y[i]);
for(i=;i<=n-;i++)
{
for(j=i+;j<=n-;j++)
{
for(k=j+;k<=n-;k++)
{
for(l=k+;l<=n;l++)
{
//printf("%d %d %d %d\n",i,j,k,l);
if(check()==)
num++;
}
}
}
} printf("Case %d: %d\n",ca++,num);
}
return ;
}
FZU 2148 Moon Game的更多相关文章
- ACM: FZU 2148 Moon Game - 海伦公式
FZU 2148 Moon Game Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64 ...
- FZU 2148 moon game (计算几何判断凸包)
Moon Game Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Submit St ...
- FZU 2148 Moon Game --判凹包
题意:给一些点,问这些点能够构成多少个凸四边形 做法: 1.直接判凸包 2.逆向思维,判凹包,不是凹包就是凸包了 怎样的四边形才是凹四边形呢?凹四边形总有一点在三个顶点的内部,假如顶点为A,B,C,D ...
- FZOJ Problem 2148 Moon Game
Proble ...
- fzu Problem 2148 Moon Game(几何 凸四多边形 叉积)
题目:http://acm.fzu.edu.cn/problem.php?pid=2148 题意:给出n个点,判断可以组成多少个凸四边形. 思路: 因为n很小,所以直接暴力,判断是否为凸四边形的方法是 ...
- FZU Problem 2148 Moon Game (判断凸四边形)
题目链接 题意 : 给你n个点,判断能形成多少个凸四边形. 思路 :如果形成凹四边形的话,说明一个点在另外三个点连成的三角形内部,这样,只要判断这个内部的点与另外三个点中每两个点相连组成的三个三角形的 ...
- 暴力(判凸四边形) FZOJ 2148 Moon Game
题目传送门 题意:给了n个点的坐标,问能有几个凸四边形 分析:数据规模小,直接暴力枚举,每次四个点判断是否会是凹四边形,条件是有一个点在另外三个点的内部,那么问题转换成判断一个点d是否在三角形abc内 ...
- Moon Game (凸四边形个数,数学题)
Problem 2148 Moon Game Accept: 24 Submit: 61 Time Limit: 1000 mSec Memory Limit : 32768 KB Pro ...
- FZU-2148-Moon Game,,几何计算~~
Problem 2148 Moon Game Time Limit: 1000 mSec Memory Limit : 32768 KB Problem Description Fat brothe ...
随机推荐
- 夺命雷公狗—angularjs—13—post参数的接收发送
我们强悍的angularjs为我们不仅仅提供了他的get接收方式,而且也有post的接收方式,我们现在做一个模拟接收后端传递过来的json的数据: <?php $arr = ['user'=&g ...
- imread函数、namedWindow函数、imshow函数、imwrite函数
1.imread函数 首先,我们看imread函数,可以在OpenCV官方文档中查到其原型如下: Mat imread(const string& filename, int flags=1 ...
- NOIP200902分数线划定
NOIP200902分数线划定 描述 世博会志愿者的选拔工作正在 A 市如火如荼的进行.为了选拔最合适的人才,A 市对所有报名的选手进行了笔试,笔试分数达到面试分数线的选手方可进入面试.面试分数线根据 ...
- oracle中的函数
ORACLE中函数 Oracle已经内建了许多函数,不同的函数有不同的作用和用法,有的函数只能作用在一个记录行上,有的能够作用在多个记录行上,不同的函数可能处理不同的数据类型.常见的 ...
- redmine plugin
http://wangsheng2008love.blog.163.com/blog/static/78201689200992064615770/
- Linux异步IO【转】
转自:http://blog.chinaunix.net/uid-24567872-id-87676.html Linux® 中最常用的输入/输出(I/O)模型是同步 I/O.在这个模型中,当请求发出 ...
- eclipse 利用已有c++代码建工程,并编译执行
如果你想建一个带Makefile的c++ 工程 1. 新建一个C++空工程,工程类型是makefile project,选择Linux GCC: 2. 将源码连同makefile文件一同作为一个文件系 ...
- 项目管理:CocoaPods建立私有仓库
CocoaPods是iOS,Mac下优秀的第三方包管理工具,类似于java的maven,给我们项目管理带来了极大的方便. 个人或公司在开发过程中,会积累很多可以复用的代码包,有些我们不想开源,又想像开 ...
- 解析八大O2O典范:他们都做了什么?
随着无线技术的发展二维码的发展以及智能手机的普及,零售的解决方案不仅在在一台电脑上解决,可以从线上到线下,为消费者贯通线上线下的购物体验.人人都爱O2O,可做得好的O2O案例却并不多.要解决利益分配. ...
- 线程属性pthread_attr_t
转:http://blog.sina.com.cn/s/blog_6dc9e4cf0100xcvk.html1. 线程属性: 使用pthread_attr_t类型表示,我 ...