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—10—angularjs里面的内置函数
我们没学一门语言或者框架,几乎里面都有各自的语法和内置函数,当然,强悍的angularjs也不例外,他的方法其实常用的没多少,因为很多都可以用源生jis几乎都能完成一大部分.. <!doctyp ...
- smarty简单介绍
smarty简单介绍 示意图如下 简单介绍smarty.class.php类的大体内容,如下: <?php class Smarty //此类就是libs中的Smarty.class.php类 ...
- 【SPFA】 最短路计数
最短路计数 [问题描述] 给出一个N个顶点M条边的无向无权图,顶点编号为1-N.问从顶点1开始,到其他每个点的最短路有几条. [输入格式] 输入第一行包含2个正整数N,M,为图的顶点数与边数. ...
- linux设备驱动归纳总结(十二):简单的数码相框【转】
本文转载自:http://blog.chinaunix.net/uid-25014876-id-116926.html linux设备驱动归纳总结(十二):简单的数码相框 xxxxxxxxxxxxxx ...
- linux设备驱动归纳总结(四):1.进程管理的相关概念【转】
本文转载自;http://blog.chinaunix.net/uid-25014876-id-64866.html linux设备驱动归纳总结(四):1.进程管理的相关概念 xxxxxxxxxxxx ...
- Web API 2 authentication with JWT
Web API 2 authentication with JWT JSON Web Token (JWT) 使用 AngularJS & NodeJS 实现基于 token 的认证应用
- Pro ASP.NET MVC 5 Framework.学习笔记.6.4.MVC的必备工具
2.5.创建链式依赖 当你请求Ninject创建一个类型,它检查该类型的依赖是否声明.它也会检查该依赖是否依赖其他类型.如果这里有附加依赖,Ninject自动解决他们,并创建请求的所有类的实例.正是由 ...
- Can't create/write to file '/tmp/#sql_887d_0.MYD' (Errcode: 17)
lsof |grep "#sql_887d_0.MYD" 如果没有被占用就可以删掉 . https://wordpress.org/support/topic/cant-creat ...
- inupt textarea提示文字(点击消失,不输入恢复)
<input name="textfield" type="text" maxlength="20" value="请输入 ...
- json-encode()怎么进行解码呢?
解决中文的一种方法就是先将中文转换为另一种编码格式,然后再使用json_encode(),最后再用解码把json串进行解码.还有一种方式就在php新版本中得到了解决,在下面的代码为展示. 以下为代码示 ...