UVa OJ 143 - Orchard Trees (果树林)

Time limit: 3.000 seconds
限时：3.000秒

 

Problem
问题

An Orchardist has planted an orchard in a rectangle with trees uniformly spaced in both directions. Thus the trees form a rectangular grid and we can consider the trees to have integer coordinates. The origin of the coordinate system is at the bottom left of the following diagram:
有一个果园主种植了一片矩形的果树林，果树之间的间隔距离在纵横两个方向都相同。想象这片果树林形成了一个矩形的格网，每棵果树的纵横坐标值均为整数。坐标系统的原点定在下图的左下角：

 

 

Consider that we now overlay a series of triangles on to this grid. The vertices of the triangle can have any real coordinates in the range 0.0 to 100.0, thus trees can have coordinates in the range 1 to 99. Two possible triangles are shown.
我们现在要在这个格网上面放上一些三角形。三角形的顶点可以是0.0到100.0之间的任意实数，这样树的横纵坐标就限定在1到99之间。上图显示了两个可能三角形。

Write a program that will determine how many trees are contained within a given triangle. For the purposes of this problem, you may assume that the trees are of point size, and that any tree (point) lying exactly on the border of a triangle is considered to be in the triangle.
写一个程序计算出在给定的一个三角形内有多少树。为简化题目，你可以认为树都一个无穷小的点，而精确位于三角形边上的树应算作在三角形内的。

 

Input and Output
输入和输出

Input will consist of a series of lines. Each line will contain 6 real numbers in the range 0.00 to 100.00 representing the coordinates of a triangle. The entire file will be terminated by a line containing 6 zeroes (0 0 0 0 0 0).
输入由多行组成，每行包含6个范围在0.00到100.00之间的整数值，分别代表三角形的三个顶点坐标的x,y。全部的输入由6个0（0 0 0 0 0 0）表示结束。

Output will consist of one line for each triangle, containing the number of trees for that triangle right justified in a field of width 4.
每个输入的三角形对应一行输出，打印出在该三角形内部的树的数量。输出的数字应右对齐至第4列。

 

Sample input
输入示例

1.5 1.5  1.5 6.8  6.8 1.5
10.7 6.9  8.5 1.5  14.5 1.5
0 0 0 0 0 0

 

Sample output
输出示例

  15
  17

 

Analysis
分析

这道题目的算法在前面已经多次处现了，就是用外积来判断点是否在三角形内。详见UVa OJ 109 - SCUD Busters (SCUD重磅炸弹)、UVa OJ 137 - Polygons (多边形)。

最直接的思路就是暴力搜索，遍例每一棵树，判定它是否在给定的多边形内。如果你的代码写的不太烂，3秒内完成绝对是没有问题的，但我们还可以进一步的优化。比如可以先判定是否在三角形的外接矩形框中。

然而这道题是可以不用遍例直接算出结果的，我已经有了一个思路，只是由于时间关系还没有实现。下面先把暴力搜索的代码贴出来（很粗糙，没有优化），等计算法的实现AC以后再更新。

 

Solution
解答

#include <algorithm>
#include <iostream>
#include <iomanip>
#include <string>
#include <vector>
struct POINT {int x; int y;};
using namespace std;
//主函数
int main(void) {
	for (POINT Coord[3], Tree; ;) { //循环处理每一组输入的数据
		POINT Min = {10000, 10000}, Max = {0, 0};
		for (int i = 0; i < 3; ++i) { //读取输入的3个顶点坐标值
			float fX, fY;
			if (0 == (cin >> fX >> fY)) {
				return 0;
			}
			Coord[i].x = (int)(fX * 100 + 0.5); //放大100倍取整
			Coord[i].y = (int)(fY * 100 + 0.5); //整数运算可保证精度
			//统计最大和最小的坐标值
			if (Coord[i].x < Min.x) {
				Min.x = Coord[i].x;
			}
			if (Coord[i].y < Min.y) {
				Min.y = Coord[i].y;
			}
			if (Coord[i].x > Max.x) {
				Max.x = Coord[i].x;
			}
			if (Coord[i].y > Max.y) {
				Max.y = Coord[i].y;
			}
		}
		//坐标最值没变，说明所给数据不在范围内
		if (Min.x > Max.x || Min.y > Max.y) {
			cout << "   0" << endl;
		}
		//全0表示输入结束
		if (Coord[0].x == 0 && Coord[0].y == 0 &&
			Coord[1].x == 0 && Coord[1].y == 0 &&
			Coord[2].x == 0 && Coord[2].y == 0) {
			break;
		}
		POINT Vec[3] = { //建立三条边的向量
			{Coord[1].x - Coord[0].x, Coord[1].y - Coord[0].y},
			{Coord[2].x - Coord[1].x, Coord[2].y - Coord[1].y},
			{Coord[0].x - Coord[2].x, Coord[0].y - Coord[2].y},
		};
		//检查三个点是否以顺时针方向给出
		if (Vec[0].x * Vec[1].y - Vec[1].x * Vec[0].y > 0) {
			swap(Coord[0], Coord[1]);
			Vec[0].x = Coord[1].x - Coord[0].x;
			Vec[0].y = Coord[1].y - Coord[0].y;
			Vec[1].x = Coord[2].x - Coord[1].x;
			Vec[1].y = Coord[2].y - Coord[1].y;
			Vec[2].x = Coord[0].x - Coord[2].x;
			Vec[2].y = Coord[0].y - Coord[2].y;
		}
		//最小值和最大值分别取整
		if (Min.x % 100 != 0) {
			Min.x = (Min.x / 100 + 1) * 100;
		}
		if (Min.y % 100 != 0) {
			Min.y = (Min.y / 100 + 1) * 100;
		}
		if (Max.x % 100 != 0) {
			Max.x = Max.x / 100 * 100;
		}
		if (Max.y % 100 != 0) {
			Max.y = Max.y / 100 * 100;
		}
		//约束最值
		if (Min.x < 100) {
			Min.x = 100;
		}
		if (Min.y < 100) {
			Min.y = 100;
		}
		if (Max.x > 9900) {
			Max.x = 9900;
		}
		if (Max.y > 9900) {
			Max.y = 9900;
		}
		int nCnt = 0, i;
		//在三角形外接矩形的范围内遍例每一个点，检查是否在三角形内
		for (Tree.y = Min.y; Tree.y <= Max.y; Tree.y += 100) {
			for (Tree.x = Min.x; Tree.x <= Max.x; Tree.x += 100) {
				for (i = 0; i < 3; ++i) {
					POINT VecT = {Coord[i].x - Tree.x, Coord[i].y - Tree.y};
					if (VecT.x * Vec[i].y - VecT.y * Vec[i].x > 0) {
						break;
					}
				}
				nCnt += (i == 3);
			}
		}
		//按格式要求输出结果
		cout << setw(4) << nCnt << endl;
	}
	return 0;
}