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
1 /*brute force solution to convex hull problem */ 2 /*only limited to there is no point on the same line with any other point*/ 3 /*use the new extreme point to find the next*/ 4 #include<stdio.h> 5 #include <stdlib.h> 6 #include <math.h> 7 #define MAX 100 8 #define MAXDISTANCE 100000 9 struct Point 10 { 11 int x; 12 int y; 13 } points[MAX]; 14 15 int main() 16 { 17 FILE *fp; 18 if((fp=fopen("ConvexHull.txt","r"))==NULL) 19 { 20 printf("Cannot open file !"); 21 exit(1); 22 } 23 int n=0; 24 int a=0; 25 int b=0; 26 int c=0; 27 int sign=0; 28 int cmp;//compare with sign 29 int point; 30 31 int count=0; 32 int start=-1; 33 int pre=-1; 34 35 while(!feof(fp)) 36 { 37 fscanf(fp,"%d,%d",&points[n].x,&points[n].y); 38 n++; 39 } 40 41 int convexHull[n]= {-1}; 42 43 44 for(int i=0; i<n; i++) 45 printf("No %d point:(%d,%d)\n",i+1,points[i].x,points[i].y); 46 47 int one=0; 48 int two=0; 49 50 51 while(one<n&&one!=start) 52 { 53 for(two=0; two<n; two++)//only calculate two=one;two<n is wrong, one‘s two can be smaller than one 54 { 55 if(one!=two&&two!=pre) 56 { 57 a=points[two].y-points[one].y; 58 b=points[one].x-points[two].x; 59 c=points[one].x*points[two].y-points[one].y*points[two].x; 60 sign=0; //whether ax+by-c is postive or not 61 for(point=0; point<n; point++) 62 if(point!=one&&point!=two) 63 { 64 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 65 66 if(sign==0) 67 sign=cmp;//if the first cmp is zero ,it can not be used for sign,thus sign will be assgined again 68 else if(sign*cmp<0) 69 break; 70 71 } 72 if(point==n)//all points on the same side 73 { 74 // printf("%d and %d are extreme points!\n",one,two); 75 if(start==-1) 76 start=one; 77 printf("(%d,%d) is extreme point!\n",points[one].x,points[one].y); 78 pre=one; 79 one=two;//not all one will ++,some one depend on two but if there is no extreme point two then one will ++ 80 /*use the new extreme point to find the next*/ 81 break; 82 } 83 } 84 }//for(two=0; two<n; two++) 85 if(two==n) 86 one++; 87 } 88 89 90 return 0; 91 }
时间: 2024-10-06 05:36:37