POJ 1265 计算几何 多边形面积 内部格点数 边上格点数

链接：

http://poj.org/problem?id=1265

题意：

给你一个多边形，求它的面积，内部格点数目，边上格点数目

题解：

pick公式：

给定顶点坐标均是整数点的简单多边形，有

　　　　　　　　　　面积=内部格点数目+边上格点数目/2+1

边界上的格点数：

把每条边当做左开右闭的区间以避免重复，一条左开右闭的线段(x1,y1)->(x2,y2)上的格点数为：

　　　　　　　　　　gcd(x2-x1,y2-y1)。

代码：

#include <map>
#include <set>
#include <cmath>
#include <queue>
#include <stack>
#include <cstdio>
#include <string>
#include <vector>
#include <cstdlib>
#include <cstring>
#include <sstream>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define mp make_pair
#define lson l,m,rt<<1  
#define rson m+1,r,rt<<1|1 
typedef long long ll;
typedef vector<int> VI;
typedef pair<int, int> PII;
const ll MOD = 1e9 + 7;
const int INF = 0x3f3f3f3f;
const int MAXN = 2e4 + 7;
// head

const double eps = 1e-8;
int cmp(double x) {
    if (fabs(x) < eps) return 0;
    if (x > 0) return 1;
    return -1;
}

const double pi = acos(-1);
inline double sqr(double x) {
    return x*x;
}
struct point {
    double x, y;
    point() {}
    point(double a, double b) :x(a), y(b) {}
    void input() {
        scanf("%lf%lf", &x, &y);
    }
    friend point operator+(const point &a, const point &b) {
        return point(a.x + b.x, a.y + b.y);
    }
    friend point operator-(const point &a, const point &b) {
        return point(a.x - b.x, a.y - b.y);
    }
    friend point operator*(const double &a, const point &b) {
        return point(a*b.x, a*b.y);
    }
    friend point operator/(const point &a, const double &b) {
        return point(a.x / b, a.y / b);
    }
    double norm() {
        return sqrt(sqr(x) + sqr(y));
    }
};
double det(point a, point b) {
    return a.x*b.y - a.y*b.x;
}
double dot(point a, point b) {
    return a.x*b.x + a.y*b.y;
}
double dist(point a, point b) {
    return (a - b).norm();
}

struct line {
    point a, b;
    line() {}
    line(point x, point y) :a(x), b(y) {}
};
double dis_point_segment(point p, point s, point t) {
    if (cmp(dot(p - s, t - s)) < 0) return (p - s).norm();
    if (cmp(dot(p - t, s - t)) < 0) return (p - t).norm();
    return fabs(det(s - p, t - p) / dist(s, t));
}
bool point_on_segment(point p, point s, point t) {
    return cmp(det(p - s, t - s)) == 0 && cmp(dot(p - s, p - t)) <= 0;
}
bool parallel(line a, line b) {
    return !cmp(det(a.a - a.b, b.a - b.b));
}
bool line_make_point(line a, line b,point &res) {
    if (parallel(a, b)) return false;
    double s1 = det(a.a - b.a, b.b - b.a);
    double s2 = det(a.b - b.a, b.b - b.a);
    res = (s1*a.b - s2*a.a) / (s1 - s2);
    return true;
}

struct polygon {
    int n;
    point a[MAXN];
    double area() {
        double sum = 0;
        a[n] = a[0];
        rep(i, 0, n) sum += det(a[i], a[i + 1]);
        return sum / 2;
    }
    point MassCenter() {
        point ans = point(0, 0);
        if (cmp(area()) == 0) return ans;
        a[n] = a[0];
        rep(i, 0, n) ans = ans + det(a[i + 1], a[i])*(a[i] + a[i + 1]);
        return ans / area() / 6;
    }
    int gcd(int a, int b) {
        return b == 0 ? a : gcd(b, a%b);
    }
    int Border_Int_Point_Num() {
        int num = 0; 
        a[n] = a[0];
        rep(i, 0, n) num += gcd(abs(int(a[i + 1].x - a[i].x)), 
            abs(int(a[i + 1].y - a[i].y)));
        return num;
    }
    int Inside_Int_Point_Num() {
        return int(area()) + 1 - Border_Int_Point_Num() / 2;
    }
};

int n;
polygon p;

int main() {
    int T;
    cin >> T;
    rep(cas, 1, T + 1) {
        cin >> n;
        p.n = n;
        rep(i, 0, n) scanf("%lf%lf", &p.a[i].x, &p.a[i].y);
        rep(i, 1, n) p.a[i] = p.a[i] + p.a[i - 1];
        int a = p.Inside_Int_Point_Num();
        int b = p.Border_Int_Point_Num();
        double c = p.area();
        printf("Scenario #%d:
%d %d %.1f

", cas, a, b, c);
    }
    return 0;
}