FZOJ2110 star(DFS)

最新推荐文章于 2016-04-02 09:33:47 发布

转载最新推荐文章于 2016-04-02 09:33:47 发布 · 61 阅读

本文介绍了一个基于星星位置判断能构成多少个内角均小于90度的锐角三角形的问题，并提供了一段C语言实现的代码。输入包括星星的数量及坐标，输出为不同锐角三角形总数。

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

Sample Output

#include<stdio.h>
#include<math.h>
typedef struct nn
{
    double x1,y1;
}node;
void cmp(double *a,double *b,double *c)
{
    double tem;
    if(*a<*b){tem=*a;*a=*b;*b=tem;}
    if(*a<*c){tem=*a;*a=*c;*c=tem;}
}
double cacreat(double x1,double y1,double x2,double y2)
{
    double edglen;
    edglen=sqrt(pow(x1-x2,2.0)+pow(y1-y2,2.0));
    return edglen;
}
int pandu(double a,double b,double c)
{
    if(b*b+c*c-a*a>0)
    return 1;
    return 0;
}
double x[105],y[105];
node s[3];
int vist[105],n,sum;
void dfs(int cout,int j)
{
    double edg1,edg2,edg3;
    int i;
    s[cout].x1=x[j];s[cout].y1=y[j];
    if(cout==2)
    {
        edg1=cacreat(s[0].x1,s[0].y1,s[1].x1,s[1].y1);
        edg2=cacreat(s[0].x1,s[0].y1,s[2].x1,s[2].y1);
        edg3=cacreat(s[1].x1,s[1].y1,s[2].x1,s[2].y1);
        cmp(&edg1,&edg2,&edg3);
        if(pandu(edg1,edg2,edg3)!=0)
                sum++;
        return ;
    }
    for(i=j+1;i<n;i++)
    if(vist[i]==0)
    {
        vist[i]=1;
        dfs(cout+1,i);
        vist[i]=0;
    }
}
int main()
{
    int i,t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i=0;i<n;i++)
        {
            scanf("%lf%lf",&x[i],&y[i]);
            vist[i]=0;
        }
        sum=0;
        for(i=0;i<n;i++)
        {
            vist[i]=1;
            dfs(0,i);
            vist[i]=0;
        }
        printf("%d\n",sum);
    }
}