2014-05-10 22:58
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 ;
}