POJ 1265 Area(Pick定理)

题目链接：POJ 1265 Area

我没听说过这定理，看别人题解做的。

Pick定理，给定顶点座标均是整点（或正方形格点）的简单多边形，皮克定理说明了其面积A和内部格点数目i、边上格点数目b的关系：A = i + b/2 - 1。

面积计算使用叉积就可以了，边上点的计算分成几种情况：如果一条线是水平或者竖直的计算很简单，如果不是这种情况，点的个数为两端点横坐标之差和纵坐标之差的gcd减去1，注意计算都是不包括端点的，端点最后加上。

#include <iostream>
#include <cstdio>
#include <cmath>

using namespace std;

const double eps = 1e-10;
const int MAX_N = 100 + 10;

struct Point
{
	int x, y;
	Point(int x=0, int y=0):x(x),y(y) { }
};

typedef Point Vector;

Vector operator + (const Vector& A, const Vector& B)
{
    return Vector(A.x+B.x, A.y+B.y);
}

Vector operator - (const Point& A, const Point& B)
{
    return Vector(A.x-B.x, A.y-B.y);
}

Vector operator * (const Vector& A, double p)
{
    return Vector(A.x*p, A.y*p);
}

bool operator < (const Point& a, const Point& b)
{
	return a.x < b.x || (a.x == b.x && a.y < b.y);
}

int dcmp(double x)
{
    if(fabs(x) < eps)
        return 0;
    else
        return x < 0 ? -1 : 1;
}

bool operator == (const Point& a, const Point &b)
{
	return dcmp(a.x-b.x) == 0 && dcmp(a.y-b.y) == 0;
}

int Cross(const Vector& A, const Vector& B)
{
    return A.x*B.y - A.y*B.x;
}

int Area2(Point A, Point B, Point C)
{
    return Cross(B - A, C - A);
}

int gcd(int a ,int b)
{
    return b == 0 ? a : gcd(b , a % b);
}
Point p[MAX_N];

int main()
{
    int T, cnt;
    cnt = 0;
    scanf("%d", &T);
    while(T--)
    {
        int n;
        scanf("%d", &n);
        p[0].x = 0, p[1].y = 0;
        int dx, dy;
        for(int i = 1; i <= n; i++)
        {
            scanf("%d%d", &dx, &dy);
            p[i].x = p[i - 1].x + dx, p[i].y = p[i - 1].y + dy;
        }
        int area = 0;
        for(int i = 1; i < n - 1; i++)
            area += Area2(p[0], p[i], p[i + 1]);
        if(area < 0)
            area = -area;
        int edge_point = 0;
        for(int i = 0; i < n; i++)
        {
            dx = abs(p[i].x - p[i + 1].x);
            dy = abs(p[i].y - p[i + 1].y);
            if(dx == 0 && dy == 0)
                continue;
            if(dx == 0)
                edge_point += dy - 1;
            else if(dy == 0)
                edge_point += dx - 1;
            else
                edge_point += gcd(dx, dy) - 1;
        }
        edge_point += n;
        printf("Scenario #%d:\n%d %d %.1lf\n\n", ++cnt, (area + 2 - edge_point) / 2, edge_point, area / 2.0);
    }
    return 0;
}