POJ 1265 Pcik定理
题意:
求多边形内部整点的个数、边上整点数、多边形的面积(点数指正点数)
题解:
[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
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
1 #include <iostream> 2 #include <cstring> 3 #include <cstdlib> 4 #include <cstdio> 5 #include <algorithm> 6 #include <cmath> 7 8 #define N 222 9 10 using namespace std; 11 //[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 12 struct PO 13 { 14 int x,y; 15 }p[N]; 16 17 int n,gs; 18 19 inline void read() 20 { 21 scanf("%d",&n); 22 for(int i=1,a=0,b=0,c,d;i<=n;i++) 23 { 24 scanf("%d%d",&c,&d); 25 a+=c; b+=d; 26 p[i].x=a; p[i].y=b; 27 } 28 } 29 30 inline int cross(PO &a,PO &b,PO &c) 31 { 32 return (b.x-a.x)*(c.y-a.y)-(c.x-a.x)*(b.y-a.y); 33 } 34 35 inline int gcd(int a,int b) 36 { 37 int ys; 38 while(b) 39 { 40 ys=a%b; 41 a=b; b=ys; 42 } 43 return a; 44 } 45 46 inline int getnum(PO &a,PO &b) 47 { 48 int x=abs(a.x-b.x),y=abs(a.y-b.y); 49 return gcd(x,y)-1; 50 } 51 52 inline int getarea() 53 { 54 int ans=0; 55 for(int i=1;i<=n;i++) ans+=cross(p[0],p[i],p[i+1]); 56 return ans; 57 } 58 59 inline void go() 60 { 61 p[n+1]=p[1]; 62 int edgenum=n; 63 for(int i=1;i<=n;i++) edgenum+=getnum(p[i],p[i+1]); 64 int area=abs(getarea()); 65 int innum=(area-edgenum)/2+1; 66 printf("Scenario #%d:\n",++gs); 67 printf("%d %d %.1lf\n\n",innum,edgenum,area*0.5); 68 } 69 70 int main() 71 { 72 int cas; scanf("%d",&cas); 73 while(cas--) read(),go(); 74 return 0; 75 }
没有人能阻止我前进的步伐,除了我自己!