zoj 1032 || poj 1265 Area 2

用pick定理求内部的整点个数：在一个平面直角坐标系内，以整点为顶点的简单多边形（任两边不交叉），它内部整点数为a，它的边上（包括顶点）的整点数为b，则它的面积S = a+b/2-1。

 

以整点为顶点的线段，覆盖的整点数为GCD(dx,dy)，其中，dxdy分别为线段横向占的点数和纵向占的点数。如果dx或dy为0，则覆盖的点数为dy或dx。（这个，不会证呐。。）

 

#include <queue>#include <stack>#include <math.h>#include <stdio.h>#include <stdlib.h>#include <iostream>#include <limits.h>#include <string.h>#include <algorithm>using namespace std;const int MAX = 110;struct point{ int x,y;};point p[MAX];double area_polygon(point p[],int n){double s1 = 0.0,s2 = 0.0;for(int i=0; i<n; i++){s1 += p[(i+1)%n].y * p[i].x;s2 += p[(i+1)%n].y * p[(i+2)%n].x;}return fabs(s1 - s2)/2.0;}int gcd(int n,int m){return m == 0 ? n : gcd(m,n%m);}int main(){int ncases,n;int dx,dy;int ind = 1;scanf("%d",&ncases);while( ncases-- ){scanf("%d",&n);scanf("%d %d",&p[0].x,&p[0].y);int pedge = 0;pedge += gcd(abs(p[0].x),abs(p[0].y));for(int i=1; i<n; i++){scanf("%d %d",&dx,&dy);pedge += gcd(abs(dx),abs(dy));p[i].x = p[i-1].x + dx;p[i].y = p[i-1].y + dy;}double area = area_polygon(p,n);int inp = (int)(area) - pedge/2 + 1;printf("Scenario #%d:/n",ind++);printf("%d %d %.1lf/n/n",inp,pedge,area);}return 0;} 

 

用通用函数写的

 

#include <queue> #include <stack> #include <math.h> #include <stdio.h> #include <stdlib.h> #include <iostream> #include <limits.h> #include <string.h> #include <algorithm> using namespace std; const int MAX = 110; struct point{ int x,y;}; point p[MAX]; double area_polygon(point p[],int n) { double s1 = 0.0,s2 = 0.0; for(int i=0; i<n; i++) { s1 += p[(i+1)%n].y * p[i].x; s2 += p[(i+1)%n].y * p[(i+2)%n].x; } return fabs(s1 - s2)/2.0; } int gcd(int n,int m) { return m == 0 ? n : gcd(m,n%m); }int intp_insegment(point a, point b){int aa = abs(b.y - a.y), bb = abs(b.x - a.x);if(aa == 0 && bb == 0)return 0;if(aa == 0)return bb - 1;if(bb == 0)return aa - 1;return gcd(aa, bb) - 1;}int intp_edge(point p[],int n){int ans = n;for(int i=0; i<n; i++)ans += intp_insegment(p[i], p[(i+1)%n]);return ans;}int main() { int ncases,n; int dx,dy; int ind = 1; scanf("%d",&ncases); while( ncases-- ) { scanf("%d",&n); scanf("%d %d",&p[0].x,&p[0].y); for(int i=1; i<n; i++) { scanf("%d %d",&dx,&dy); p[i].x = p[i-1].x + dx; p[i].y = p[i-1].y + dy; }int pedge = intp_edge(p,n); double area = area_polygon(p,n); int inp = (int)(area) - pedge/2 + 1; printf("Scenario #%d:/n",ind++); printf("%d %d %.1lf/n/n",inp,pedge,area); } return 0; }