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 ",maxz); 72 } 73 return 0; 74 }