Careercup - Microsoft面试题 - 5120588943196160

2014-05-10 22:58

题目链接

原题：

Three points are given A(x1, y1), B(x2, y2), C(x3, y3). Write a method returning an array of points (x, y) inside the triangle ABC.

题目：给定三个点，找出所有这三点组成的三角形内的整点。(不能组成三角形也无所谓，结果为空即可。)

解法：出题者没有说是整点，但如果不是整点，就有无穷多个了。求整点的个数可以用Pick定律的公式。要求出所有整点的话，我的方法是找出一个内部的整点，然后向四个方向进行DFS，直到找出所有点。搜索过程中，需要判断点是否在三角形内部。我的判断方法是计算面积。对于整数问题，计算公式不要引入浮点数，误差是不必要的。所以海伦公式不可行，用行列式计算面积更为方便、准确。

代码：

 // http://www.careercup.com/question?id=5120588943196160
 #include <iostream>
 #include <unordered_set>
 #include <vector>
 using namespace std;

 struct Point {
     int x;
     int y;
     Point(int _x = , int _y = ): x(_x), y(_y) {};
 };

 struct hashFunctor {
     size_t operator () (const Point &p) {
         return p.x *  + p.y;
     };
 };

 struct equalFunctor {
     bool operator () (const Point &p1, const Point &p2) {
         return p1.x == p2.x && p1.y == p2.y;
     };
 };

 typedef unordered_set<Point, hashFunctor, equalFunctor> point_set;

 int twoArea(int x1, int y1, int x2, int y2, int x3, int y3)
 {
     return abs(x1 * (y2 - y3) + x2 * (y3 - y1) + x3 * (y1 - y2));
 }

 bool inside(int x[], int y[], int px, int py)
 {
     int sum = ;

     sum += twoArea(x[], y[], x[], y[], px, py);
     sum += twoArea(x[], y[], x[], y[], px, py);
     sum += twoArea(x[], y[], x[], y[], px, py);
     return sum == twoArea(x[], y[], x[], y[], x[], y[]);
 }

 void DFS(int x[], int y[], int px, int py, point_set &um, point_set &visited)
 {
     static const int dir[][] = {
         {-, },
         {+, },
         {, -},
         {, +}
     };
     int i;
     int newx, newy;

     visited.insert(Point(px, py));
     um.insert(Point(px, py));

     for (i = ; i < ; ++i) {
         newx = px + dir[i][];
         newy = py + dir[i][];
         if (visited.find(Point(newx, newy)) == visited.end() &&
             inside(x, y, newx, newy)) {
             DFS(x, y, newx, newy, um, visited);
         }
     }
 }

 void insidePoints(int x[], int y[], vector<Point> &points)
 {
     point_set um;
     point_set visited;
     int mx, my;

     mx = (x[] + x[] + x[]) / ;
     my = (y[] + y[] + y[]) / ;
     DFS(x, y, mx, my, um, visited);

     point_set::const_iterator usit;
     for (usit = um.begin(); usit != um.end(); ++usit) {
         points.push_back(Point(usit->x, usit->y));
     }
     um.clear();
     visited.clear();
 }

 int main()
 {
     int x[];
     int y[];
     vector<Point> points;
     int i;
     int n;

     while (cin >> x[] >> y[]) {
         cin >> x[] >> y[];
         cin >> x[] >> y[];
         insidePoints(x, y, points);
         n = (int)points.size();
         for (i = ; i < n; ++i) {
             cout << points[i].x << ' ' << points[i].y << endl;
         }
         points.clear();
     }

     return ;
 }