poj 1265 Pick定理(模版题)

题意：在网格上给出一个机器人监控的行走路线，开始是从(0,0)出发，每一次在前一步的基础上
向右走dx步，向上走dy步，最终走回起点，求路线形成的多边形的边上、内部的点的个数以及该
多边形的面积

题解：pick定理(模板题)

#include <iostream>

using namespace std;

struct point
{
int x,y;
};

point points[105];

int gcd(int a,int b)
{
return b?gcd(b,a%b):a;
}

//多边形上的网格点个数
int grid_onedge(int n,point* p)
{
int i,ret=0;
for (i=0;i<n;i++)
ret+=gcd(abs(p[i].x-p[(i+1)%n].x),abs(p[i].y-p[(i+1)%n].y));
return ret;
}

//多边形内的网格点个数
int grid_inside(int n,point* p)
{
int i,ret=0;
for (i=0;i<n;i++)
ret+=p[(i+1)%n].y*(p[i].x-p[(i+2)%n].x);
return (abs(ret)-grid_onedge(n,p))/2+1;
}

int main(void)
{
int t,cas;
cin >> t;
for(cas=1; cas<=t; cas++)
{
int m,dx,dy;
cin >> m;
cin >> points[0].x >> points[0].y;
for(int i=1; i<m; i++)
{
cin >> dx >> dy;
points[i].x = points[i-1].x + dx;
points[i].y = points[i-1].y + dy;
}

int ipt,ept;
double area;
ipt = grid_inside(m,points);
ept = grid_onedge(m,points);
area = ipt + ept/2.0 - 1;
cout << "Scenario #" << cas <<  ":" << endl;
printf("%d %d %.1lf\n\n",ipt,ept,area);
}
return 0;
}