UVA 10088 Trees on My Island

又是这种题哈，计算多边形里面内整点的个数。

 

以前做过，这次直接上模板了。。

 

 

[Pick定理]    设以整数点为顶点的多边形的面积为S， 多边形内部的整数点数为N, 多边形边界上的整数点数为L， 则 

N + L/2 - 1 = S

 

注意范围，我用long long 了，因为中间乘法往转换了 WA了一次。。。T T 

 

#include <queue> #include <stack> #include <math.h> #include <stdio.h> #include <stdlib.h> #include <iostream> #include <limits.h> #include <string.h> #include <string> #include <algorithm> using namespace std; const int MAX = 1010; struct point{ int x,y;}; point p[MAX]; 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; } long long intp_edge(point p[],int n) { long long ans = n; for(int i=0; i<n; i++) ans += intp_insegment(p[i], p[(i+1)%n]); return ans; } double area_polygon(point p[],int n) { double s = 0.0; for(int i=0; i<n; i++) s += p[(i+1)%n].y *1ll* p[i].x - p[(i+1)%n].x*1ll * p[i].y; return fabs(s)/2.0; } long long intp_inpolygon(point p[],int n) { double area = area_polygon(p,n); long long pinedge = intp_edge(p,n); return (long long)(area) - pinedge/2ll + 1; } int main() { int n; while( ~scanf("%d",&n) && n ) { for(int i=0; i<n; i++) scanf("%d %d",&p[i].x,&p[i].y); long long ans = intp_inpolygon(p,n); printf("%lld/n",ans); } return 0; }