poj 1265 ||poj2954 pick公式 网格

/ :求一个多边形中在网格内点的个数，在边上的点的个数，多边形的面积
//poj2954:求三角形内整点个数 ，两题大同小异，

//利用pick公式：面积=内点+边上的点/2-1；

代码1：poj1265

#include<iostream>#include<cstdio>#include<math.h>#define abs(x)(x>=0? x:-x)int n;int gcd(int a,int b){return b? gcd(b,a%b):a;}int area(int x1,int y1,int x2,int y2){    return (x1*y2-x2*y1);}int main(){int cas;scanf("%d",&cas);int k,i;for(k=1;k<=cas;k++){scanf("%d",&n);int in=0;int on=0;int x1=0,y1=0,x2,y2;double res=0;for(i=0;i<n;i++){int x,y;scanf("%d%d",&x,&y);on+=gcd(abs(x),abs(y));x2=x1+x;y2=y1+y;res+=area(x1,y1,x2,y2);x1=x2;y1=y2;}printf("Scenario #%d:\n",k);res/=2.0;in=fabs(res)+1-on/2.0;printf("%d %d %0.1lf\n\n",in,on,res);}return 0;}

代码2：2954

#include<iostream>#include<cstdio>#include<math.h>int gcd(int a,int b){return b? gcd(b,a%b): a;}struct point {int x,y;};point p[3];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;}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(){while(scanf("%d%d%d%d%d%d",&p[0].x,&p[0].y,&p[1].x,&p[1].y,&p[2].x,&p[2].y),p[0].x||p[0].y||p[1].x||p[1].y||p[2].x||p[2].y){printf("%d\n",grid_inside(3,p));}return 0;}