POJ 2954 Triangle(计算几何---Pick定理)

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原文链接：http://www.cnblogs.com/lidaojian/archive/2012/05/06/2486491.html

Triangle

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3627		Accepted: 1586

Description

A lattice point is an ordered pair (x, y) where x and y are both integers. Given the coordinates of the vertices of a triangle (which happen to be lattice points), you are to count the number of lattice points which lie completely inside of the triangle (points on the edges or vertices of the triangle do not count).

Input

The input test file will contain multiple test cases. Each input test case consists of six integers x₁, y₁, x₂, y₂, x₃, and y₃, where (x₁, y₁), (x₂, y₂), and (x₃, y₃) are the coordinates of vertices of the triangle. All triangles in the input will be non-degenerate (will have positive area), and −15000 ≤ x₁, y₁, x₂, y₂, x₃, y₃ ≤ 15000. The end-of-file is marked by a test case with x₁ = y₁ = x₂ = y₂ = x₃ = y₃ = 0 and should not be processed.

Output

For each input case, the program should print the number of internal lattice points on a single line.

Sample Input

0 0 1 0 0 1

0 0 5 0 0 5

0 0 0 0 0 0

Sample Output

Source

Stanford Local 2004

解题报告：题意就是求三角形内部有多少个点（整数坐标点），利用Pick定理可以求出，多边形的面积、内部的点、及边上的点的关系满足area = in + on / 2 - 1，在内部的点的个数in = area - on / 2 + 1；题目和POJ1265相同，下面写了两种求面积的代码

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
using namespace std;
int on, in, area;
struct Point
{
    double x;
    double y;
    int flag;
}p[4];
int Gcd(int a, int b)//求最大公约数
{
    int temp;
    if (a < b)
    {
        temp = a;
        a = b;
        b = temp;
    }
    if (b == 0)
    {
        return a;
    }
    return Gcd(b, a % b);
}
int Abs(int a)//求绝对值
{
    if (a < 0)
    {
        return -a;
    }
    return a;
}
/*double Multi(Point p1, Point p2, Point p3)//叉乘
{
    return (p1.x - p3.x) * (p2.y - p3.y) - (p2.x - p3.x) * (p1.y - p3.y);
}*/
double Cross(Point p1, Point p2)//叉积
{
    return p1.x * p2.y - p1.y * p2.x;
}
int main()
{
    int i;
    double xx, yy;
    memset(p, 0, sizeof(p));
    while (scanf("%lf%lf", &p[0].x, &p[0].y)!= EOF)
    {
        if (p[0].x == 0 && p[0].y == 0)
        {
            p[0].flag = 1;
        }
        for (i = 1; i < 3; ++i)
        {
            scanf("%lf%lf", &p[i].x, &p[i].y);
            if (p[i].x == 0 && p[i].y == 0)
            {
                p[i].flag = 1;
            }
        }
        if (p[0].flag == 1 && p[1].flag == 1 && p[2].flag == 1)
        {
            break;
        }
        on = 0;
        area = 0;
        for (i = 0; i < 3; ++i)
        {
            xx = p[i].x - p[(i + 1) % 3].x;
            yy = p[i].y - p[(i + 1) % 3].y;
            on += Gcd(Abs(xx), Abs(yy));
            area += Cross(p[i], p[(i + 1) % 3]);
        }
        area = Abs(area) / 2;
        //area = Abs(Multi(p[0], p[1], p[2])) / 2;
        in = (int)(area - on / 2 + 1);//利用Pick定理
        printf("%d\n", in);
    }
    return 0;
}

转载于:https://www.cnblogs.com/lidaojian/archive/2012/05/06/2486491.html