题意:求给定的多边形的内部点的个数,边上的点的个数。及面积
思路:多边形内的坐标点的个数:PICK公式:S=I+E/2-1 S,面积,I多边形的内坐标点的个数,E多边形的边上坐标点的个数。
线段的端点在坐标点上,其经过的坐标点的个数:GCD(dx,dy)+1;
#include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <iostream> #include <queue> #include <stack> using namespace std; const double EPS = 1e-12; const double INF = 1e20; bool zero(double t){return -EPS<t&&t<EPS;} struct cvector{ double x,y; cvector(){} cvector(double a,double b){x=a,y=b;} }; cvector operator-(cvector a,cvector b){ return cvector(a.x-b.x,a.y-b.y); } cvector operator+(cvector a,cvector b){ return cvector(a.x+b.x,a.y+b.y); } cvector operator*(double k,cvector a){ return cvector(k*a.x,k*a.y); } double operator*(cvector a,cvector b){ return a.x*b.x+a.y*b.y; } double operator^(cvector a,cvector b){ return a.x*b.y-b.x*a.y; } double length(double t){return t<0?-t:t;} double length(cvector t){return sqrt(t*t);} struct cpoint{ int x,y; void get(){scanf("%d%d",&x,&y);} }; cvector operator-(cpoint a,cpoint b){ return cvector(a.x-b.x,a.y-b.y); } double dist(cpoint a,cpoint b){ return length(a-b); } int n; cpoint re[109]; double x_mul(cpoint a,cpoint b,cpoint c){ return (a-c)^(b-c); } int gcd(int x,int y){ return x==0?y:gcd(y%x,x); } int main() { freopen("in.txt","r",stdin); int cas,T = 1; scanf("%d",&cas); while(cas--) { if(T>1) printf("\n"); scanf("%d",&n); re[0].get(); for(int i=1;i<n;i++) re[i].get(),re[i].x+=re[i-1].x,re[i].y+=re[i-1].y; re[n] = re[0]; double area = 0; cpoint o;o.x=0,o.y=0; for(int i=0;i<n;i++) { area += (x_mul(re[i],re[i+1],o)/2); } if(area<0) area = -area; int a=0,b=0; for(int i=0;i<n;i++) { int x = abs(re[i].x-re[i+1].x); int y = abs(re[i].y-re[i+1].y); a+= gcd(x,y); } b = area - a/2+1; printf("Scenario #%d:\n",T++); printf("%d %d %.1lf\n",b,a,area); } return 0; }