poj2780Linearity(多点共线）

链接

判断最多多少点在一条直线上，

可以枚举每一个点为坐标系的原点，其它点变成相应的位置，然后求得过原点及其点的斜率，排序找一下最多相同的。

 1 #include <iostream>

 2 #include<cstdio>

 3 #include<cstring>

 4 #include<algorithm>

 5 #include<stdlib.h>

 6 #include<vector>

 7 #include<cmath>

 8 #include<queue>

 9 #include<set>

10 #include<map>

11 using namespace std;

12 #define N 1010

13 #define LL long long

14 #define INF 0xfffffff

15 const double eps = 1e-8;

16 const double pi = acos(-1.0);

17 const double inf = ~0u>>2;

18 

19 struct point

20 {

21     int x,y;

22     point(int x=0,int y=0):x(x),y(y){}

23 }p[N];

24 double o[N];

25 

26 typedef point pointt;

27 pointt operator -(point a,point b)

28 {

29     return point(a.x-b.x,a.y-b.y);

30 }

31 int dcmp(double x)

32 {

33     if(fabs(x)<eps) return x;

34     return x<0?-1:1;

35 }

36 

37 int main()

38 {

39     int n,i,j;

40     while(scanf("%d",&n)!=EOF)

41     {

42         for(i = 1; i <= n; i++)

43         {

44             scanf("%d%d",&p[i].x,&p[i].y);

45         }

46         int maxz=0;

47         for(i = 1; i <= n; i++)

48         {

49             int ans = 0 ,g=0;

50             for(j = 1; j <= n ;j++)

51             {

52                 int x = p[j].x-p[i].x;

53                 int y = p[j].y-p[i].y;

54                 if(x==0)

55                 ans++;

56                 else o[g++] = y*1.0/x;

57             }

58             sort(o,o+g);

59             int num = 2;

60             for(j = 1; j < g; j++)

61             if(dcmp(o[j]-o[j-1])==0)

62             num++;

63             else

64             {ans = max(ans,num);

65                 num = 2;

66 

67             }

68             ans = max(ans,num);

69             maxz = max(maxz,ans);

70         }

71         printf("%d\n",maxz);

72     }

73     return 0;

74 }

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