HDU 1432 lining up ()

Http://acm.hdu.edu.cn/showproblem.php? PID = 1, 1432

 

Question:

Given N points on a two-dimensional plane, calculate the maximum number of points that a straight line can pass through.

 

Solution:

Because the N (1 ~ 700), so enumeration, the time complexity is O (n ^ 3), does not time out.

Enumerate two vertices and determine whether the remaining vertices are in this straight line.

 

AC code:

 1 #include<cstdio> 2  3 struct Point{ 4     int x, y; 5  6     Point(int x = 0, int y = 0): x(x), y(y){} 7  8     void scan(){ 9         scanf("%d%d", &x, &y);10     }11 };12 13 typedef Point Vector;14 15 Vector operator - (Vector A, Vector B){//重载结构体减号16     return Vector(A.x - B.x, A.y - B.y);17 }18 19 int cross(Vector A, Vector B){//叉乘20     return A.x * B.y - A.y * B.x;21 }22 23 Point p[700];24 int n;25 26 int main(){27     while(~scanf("%d", &n)){28         for(int i = 0; i < n; ++i){29             p[i].scan();30         }31 32         int best = 0;//记录直线穿过最多的点个数33         for(int i = 0; i < n; ++i){34             for(int j = i + 1; j < n; ++j){35                 int num = 2;//因为直线穿过i和j点，所以直线上已经有两个点了36                 for(int k = j + 1; k < n; ++k){37                     if(!cross(p[i] - p[j], p[i] - p[k])){//判断点k在直线ij上38                         ++num;39                     }40                 }41                 if(best < num){//更新最优值 42                     best = num;43                 }44             }45         }46         printf("%d\n", best);47     }48     return 0;49 }

 

HDU 1432 lining up ()