poj 2954

最新推荐文章于 2022-07-08 20:06:12 发布

原创最新推荐文章于 2022-07-08 20:06:12 发布 · 931 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#input #struct #each #output #pair #file

ACM_计算几何专栏收录该内容

6 篇文章

订阅专栏

本文介绍了一个算法，用于计算给定三角形内部（不包括边界）的整点数量。输入为三角形三个顶点的坐标，通过计算面积及边界上的整点数来确定内部整点数。

Triangle

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2866		Accepted: 1218

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

0
6

Source

Stanford Local 2004

#include <stdio.h>
#include <stdlib.h>
typedef struct TPoint
{
    int x;
    int y;
} TPoint;
typedef struct TLine
{
    int a, b, c;
} TLine;
int triangleArea(TPoint p1, TPoint p2, TPoint p3)
{
    //已知三角形三个顶点的坐标，求三角形的面积
    int k = p1.x * p2.y + p2.x * p3.y + p3.x * p1.y
            - p2.x * p1.y - p3.x * p2.y - p1.x * p3.y;
    if(k < 0) return -k;
    else return k;
}
TLine lineFromSegment(TPoint p1, TPoint p2)
{
    //线段所在直线,返回直线方程的三个系统
    TLine tmp;
    tmp.a = p2.y - p1.y;
    tmp.b = p1.x - p2.x;
    tmp.c = p2.x * p1.y - p1.x * p2.y;
    return tmp;
}
void swap(int &a, int &b)
{
    int t;
    t = a;
    a = b;
    b = t;
}
int Count(TPoint p1, TPoint p2)
{
    int i, sum = 0, y;
    TLine l1 =  lineFromSegment(p1, p2);
    if(l1.b == 0) return abs(p2.y - p1.y) + 1;
    if(p1.x > p2.x) swap(p1.x, p2.x); //这里没有交换WA两次
    for(i = p1.x; i <= p2.x; i++)
    {
        y = -l1.c - l1.a * i;
        if(y % l1.b == 0) sum++;
    }
    return sum;
}
int main()
{
    TPoint p1, p2, p3;
    while(scanf("%d%d%d%d%d%d", &p1.x, &p1.y, &p2.x, &p2.y, &p3.x, &p3.y) != EOF)
    {
        if(p1.x == 0 && p1.y == 0 && p2.x == 0 && p2.y == 0 && p3.x == 0 && p3.y == 0) break;
        int  A = triangleArea(p1, p2, p3);//A为面积的两倍
        int b = 0;
        int i;
        b = Count(p1, p2) + Count(p1, p3) + Count(p3, p2) - 3;//3个顶点多各多加了一次
        //i = A / 2- b / 2 + 1;
        i = (A - b) / 2 + 1;
        printf("%d/n", i);
    }
    return 0;
}