ACM--计算几何--FZU--2110--Star

题目oj地址：http://acm.fzu.edu.cn/problem.php?pid=2110

                                                   Problem 2110 Star

                                                                                            Accept: 852    Submit: 2543
                                                                       Time Limit: 1000 mSec    Memory Limit : 32768 KB

Problem Description

Overpower often go to the playground with classmates. They play and chat on the playground. One day, there are a lot of stars in the sky. Suddenly, one of Overpower’s classmates ask him: “How many acute triangles whose inner angles are less than 90 degrees (regarding stars as points) can be found? Assuming all the stars are in the same plane”. Please help him to solve this problem.

Input

The first line of the input contains an integer T (T≤10), indicating the number of test cases.

For each test case:

The first line contains one integer n (1≤n≤100), the number of stars.

The next n lines each contains two integers x and y (0≤|x|, |y|≤1,000,000) indicate the points, all the points are distinct.

Output

For each test case, output an integer indicating the total number of different acute triangles.

Sample Input

130 010 05 1000

Sample Output

1

=================================傲娇的分割线============================

题意：给你一些点，计算这些点能组成多少个锐角三角形。

       此题最重要的判断条件是锐角三角形计算公式：a*a+b*b>c*c (两边之间的距离大于第三边）

      直接使用暴力破解，进行三次循环

<pre name="code" class="cpp">#include <iostream>
using namespace std;
typedef struct Point{
   double x,y;

}Point;
Point point[110];
/**
   此题最重要的判断条件是
   锐角三角形计算公式：a*a+b*b>c*c (两边之间的距离大于第三边）
   直接使用暴力破解，进行三次循环
*/
int main(){
   int n;
   cin>>n;
   int i,j,k;
   while(n--){
     int m;
     cin>>m;
     int result=0;
     double a,b,c;//构成三角形的三条边的长度
     //将所有的点录入进point数组中
     for(i=0;i<m;i++){
        cin>>point[i].x;
        cin>>point[i].y;
     }
     //三个循环
     for(i=0;i<m;i++){
        for(j=i+1;j<m;j++){
            for(k=j+1;k<m;k++){
                a=(point[i].x-point[j].x)*(point[i].x-point[j].x)+
                (point[i].y-point[j].y)*(point[i].y-point[j].y);//计算两点之间的距离

               b=(point[i].x-point[k].x)*(point[i].x-point[k].x)+
                (point[i].y-point[k].y)*(point[i].y-point[k].y);//计算两点之间的距离

                c=(point[j].x-point[k].x)*(point[j].x-point[k].x)+
                (point[j].y-point[k].y)*(point[j].y-point[k].y);//计算两点之间的距离

                if(a+b>c&&a+c>b&&b+c>a){//两边之和大于第三边
                    result++;
                }
            }

        }
     }
     cout<<result<<endl;
   }
   return 0;
}

参考博客：http://blog.csdn.net/whjkm/article/details/45897521