POJ 3335 Rotating Scoreboard(多边形的核)

题目链接

我看的这里：http://www.cnblogs.com/ka200812/archive/2012/01/20/2328316.html

然后整理一下当做模版。0换成eps，会wa，应该要写成精度特判把。

 #include <cstdio>
 #include <cstring>
 #include <string>
 #include <cmath>
 #include <algorithm>
 using namespace std;
 #define eps 1e-8
 #define N 110
 struct point
 {
     double x,y;
 }p[N],temp[N],pre[N];
 int n,m;
 double a,b,c;
 void getline(point x,point y)
 {
     a = y.y - x.y;
     b = x.x - y.x;
     c = y.x * x.y - x.x * y.y;
 }
 point intersect(point x,point y) //获取直线ax+by+c==0  和点x和y所连直线的交点
 {
     double u = fabs(a*x.x+b*x.y+c);
     double v = fabs(a*y.x+b*y.y+c);
     point ans;
     ans.x = (x.x*v+y.x*u)/(u+v);
     ans.y = (x.y*v+y.y*u)/(u+v);
     return ans;
 }
 void cut()
 {
     int num = ,i;
     for(i = ;i <= m;i ++)
     {
         if(a*p[i].x + b*p[i].y + c >= )
         {
             temp[++num] = p[i];
         }
         else
         {
             if(a*p[i-].x + b*p[i-].y + c > )
             temp[++num] = intersect(p[i],p[i-]);
             if(a*p[i+].x + b*p[i+].y + c > )
             temp[++num] = intersect(p[i],p[i+]);
         }
     }
     for(i = ;i <= num;i ++)
     p[i] = temp[i];
     p[] = p[num];
     p[num+] = p[];
     m = num;
 }
 void fun()
 {
     int i;
     m = n;
     for(i = ;i <= n;i ++)
     {
         getline(pre[i],pre[i+]);
         cut();
     }
 }
 int main()
 {
     int i,t;
     scanf("%d",&t);
     while(t--)
     {
         scanf("%d",&n);
         for(i = ;i <= n;i ++)
         {
             scanf("%lf%lf",&pre[i].x,&pre[i].y);
             p[i] = pre[i];
         }
         pre[n+] = pre[];
         p[n+] = p[];
         p[] = p[n];
         fun();
         if(m)
         printf("YES\n");
         else
         printf("NO\n");
     }
     return ;
 }