暴力法解凸包

给定平面上一系列的点，用暴力法求解它们的凸包，此算法比普通的暴力法要优化，用新找到的极点去寻找下一个极点。此算法不能用于任何两个点在一直线上的情况。

输入 ConvexHull.txt

7,8
10,17
14,14
15,23
16,12
17,3
22,17
24,4
26,18

C代码

/*brute force solution to convex hull problem  */
/*only limited to there is no point on the same line with any other point*/
/*use the new extreme point to find the next*/
#include<stdio.h>
#include <stdlib.h>
#include <math.h>
#define MAX 100
#define MAXDISTANCE 100000
struct Point
{
    int x;
    int y;
} points[MAX];

int main()
{
    FILE *fp;
    if((fp=fopen("ConvexHull.txt","r"))==NULL)
    {
        printf("Cannot open file !");
        exit(1);
    }
    int n=0;
    int a=0;
    int b=0;
    int c=0;
    int sign=0;
    int cmp;//compare with sign
    int point;
    
    int count=0;
    int start=-1;
    int pre=-1;

    while(!feof(fp))
    {
        fscanf(fp,"%d,%d",&points[n].x,&points[n].y);
        n++;
    }

    int convexHull[n]= {-1};


    for(int i=0; i<n; i++)
        printf("No %d point:(%d,%d)
",i+1,points[i].x,points[i].y);

    int one=0;
    int two=0;


    while(one<n&&one!=start)
    {    
        for(two=0; two<n; two++)//only calculate two=one;two<n is wrong, one's two can be smaller than one 
        {
            if(one!=two&&two!=pre)
            {
                    a=points[two].y-points[one].y;
                    b=points[one].x-points[two].x;
                    c=points[one].x*points[two].y-points[one].y*points[two].x;
                    sign=0;   //whether ax+by-c is postive or not
                    for(point=0; point<n; point++)
                        if(point!=one&&point!=two)
                        {
                            cmp=a*points[point].x+b*points[point].y-c; //cmp can be zero, if cmp is zero it's on the line,one and two are can be still extreme points
                        
                            if(sign==0)
                                sign=cmp;//if the first cmp is zero ,it can not be used for sign,thus sign will be assgined again
                            else if(sign*cmp<0)
                                break;

                        }
                    if(point==n)//all points on the same side
                    {
                        //    printf("%d and %d are extreme points!
",one,two);
                        if(start==-1)
                            start=one;
                        printf("(%d,%d) is extreme point!
",points[one].x,points[one].y);
                        pre=one;
                        one=two;//not all one will ++,some one depend on two but if there is no extreme point two then one will ++ 
                        /*use the new extreme point to find the next*/
                        break;
                    }                
            }
        }//for(two=0; two<n; two++)
        if(two==n)
            one++;
    }


    return 0;
}