1054

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;

#define N  5003
#define prim 10079

int r,c,n;

struct point
{
 int x,y;
 int next;// point to next point's number
 bool operator<(const struct point &u) const
 {
   if(y == u.y)
     return x   return y }
};
int dp[N][N]; //存放j延伸到i射线上点的个数
int hash[prim];//hash 链表
point pn[N];
bool inner(int x,int y) //是否在田地范围内
{
 return (x>=1 && x<=r && y>=1 && y<=c);
}
int hash_func(int x,int y) //hash 函数，这里完全是要这个函数优劣了
{
 return (x+y+x*y)%prim;
}

int hash_find(int x,int y) //有这样的需求
{
  int h = hash_func(x,y);
  int j = hash[h];
  while(j!=-1)
  {
   if(x == pn[j].x && y == pn[j].y)
     return j;
   j = pn[j].next;
  }
  return -1;
}

void hash_insert(const struct point & p,int i)
{
 int h= hash_func(p.x,p.y);
 pn[i].next = hash[h];
 hash[h]=i;
}

int main()
{

 #ifdef LOCAL
 freopen("data.in","r",stdin);
 freopen("data.out","w",stdout);
 #endif
 
 cin>>r>>c;
 cin>>n;
 memset(hash,0xff,sizeof(hash)); //all FFFFF
 int i,j;
 for(i=0;i  scanf("%d %d",&pn[i].x,&pn[i].y);
  
 sort(pn,pn+n); //为dp 规定一个顺序，状态转移必须，明确位置关系
 
 for(i=0;i  printf("%d %d\t",pn[i].x,pn[i].y);
 cout< int ha;
 for(i=0;i  hash_insert(pn[i],i);
 int x0,y0,x1,y1,maxp = 2;
 for(i=1;i {
   for(j=0;j   {
     x0 = 2*pn[j].x - pn[i].x;  //pn[j].x<=pn[i].x,因为排过序了
     y0 = 2*pn[j].y - pn[i].y;
     if(inner(x0,y0))
     {
       ha = hash_find(x0,y0); //找点号
       if( ha !=-1 && dp[j][ha]!=-1)
         dp[i][j] = dp[j][ha] +1; //转移很巧妙,要点几何知识,
       else
         dp[i][j] = -1;    //此条射线以后也无法满足要求，能进必然能出，否则不是合法轨迹
     }
     else dp[i][j]=2;
     
     x1 = 2*pn[i].x - pn[j].x;
     y1 = 2*pn[i].y - pn[j].y;
     if(!inner(x1,y1))   //不可以再向右延伸了，记录最大数
     {
      maxp = max(maxp,dp[i][j]);
      printf("dp[%d][%d]=%d\n",i,j,dp[i][j]);
     }
   }
 }
 if(maxp>2)
   cout< else
   cout<<0;
 return 0;
}