POJ 2954(Pick公式)

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Triangle

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 4203 Accepted: 1858

Description

求出一个已知3点坐标的格点三角形的里面(边上的不算)有多少点.

Input

多组数据。每一行输入x₁, y₁, x₂, y₂, x₃,y₃, 表示格点三角形 (x₁, y₁), (x₂, y₂), (x₃, y₃), −15000 ≤ x₁, y₁, x₂, y₂, x₃, y₃ ≤ 15000. 数据以6个0结尾.

Output

每组数据输出一行，表示格点三角形里面的点数.

Sample Input

0 0 1 0 0 10 0 5 0 0 50 0 0 0 0 0

Sample Output

06

Source

Stanford Local 2004

Pick定理：S=I+E/2-1 (E表示格点多边形边上的点,I表示格点多边形内的点)

Pick推论：格点多边形S=0.5k (k是正整数)

边上格点数公式：线段(x1,y1)-(x2,y2)的格点数=gcd(abs(x1-x2),abs(y1-y2))+1

#include<cstdio>#include<cstdlib>#include<cstring>#include<cmath>#include<iostream>#include<algorithm>#include<functional>using namespace std;#define MAXX (15000)int n;int gcd(int a,int b){if (a<b) swap(a,b);if (b==0) return a;else return gcd(b,a%b);}struct P{int x,y;P(){}P(int _x,int _y):x(_x),y(_y){}friend istream& operator>>(istream& cin,P &a){cin>>a.x>>a.y;return cin;}friend bool operator||(bool b,P &a){return b||a.x||a.y;}}a[3];struct S{P s,t;S(){}S(P _s,P _t):s(_s),t(_t){}friend bool operator||(bool b,S &a){return b||a.s||a.t;}friend bool operator||(S &a,S &b){return 0||a||b;}int dx(){return abs(t.x-s.x);}int dy(){return abs(t.y-s.y);}int E(){return gcd(dx(),dy())+1;}};struct T{S c[3];S& operator[](int i){return c[i];}friend istream& operator>>(istream& cin,T &c){cin>>a[0]>>a[1]>>a[2];c[0]=S(a[0],a[1]);c[1]=S(a[1],a[2]);c[2]=S(a[2],a[0]);return cin;}friend bool operator&&(bool b,T &a){return b&&(a[0]||a[1]||a[2]);}int area2(){return c[0].s.x*c[1].s.y+c[1].s.x*c[2].s.y+c[2].s.x*c[0].s.y-c[1].s.x*c[0].s.y-c[2].s.x*c[1].s.y-c[0].s.x*c[2].s.y;}int E(){return c[0].E()+c[1].E()+c[2].E()-3;}double _S(){return (double)abs(area2())/2;}int I(){return _S()-E()/2+1;}}c;int main(){while (cin>>c&&c){cout<<c.I()<<endl;}return 0;}