poj1054

【题意】

有一个r*c的方格，青蛙会以相同的向量v=(x,y)跳过，跳过的地方留下痕迹，青蛙在格子外哪里出发都有可能，问有最多有多少只青蛙。

【输入】

r c

n

以下n行每行描述一个点。

【输出】

最多有多少只青蛙。

比较暴力……

program poj1054;type  point=record          x,y:longint;        end;var  n,r,c,i,j,k,ans,x,y:longint;  v:array [0..5001,0..5001] of boolean;  p:array [0..5001] of point;function max (a,b:longint):longint;begin  if a>b then exit(a)         else exit(b);end;procedure swap (var a,b:point);var  i:point;begin  i:=a;  a:=b;  b:=i;end;procedure qsort (s,e:longint);var  i,j:longint;  k:point;begin  if s>=e then exit;  i:=s;  j:=e;  k:=p[(s+e) div 2];  while i<=j do    begin      while (p[i].x<k.x)or((p[i].x=k.x)and(p[i].y<k.y)) do inc(i);      while (p[j].x>k.x)or((p[j].x=k.x)and(p[j].y>k.y)) do dec(j);      if i>j then break;      swap(p[i],p[j]);      inc(i);      dec(j);    end;  qsort(s,j);  qsort(i,e);end;function dfs (x,y,xx,yy:longint):longint;var  k:longint;begin  k:=2;  while (x+xx>0)and(x+xx<=r)and(y+yy>0)and(y+yy<=c) do    if v[x+xx,y+yy] then      begin        inc(k);        x:=x+xx;        y:=y+yy;      end                    else      exit(0);  exit(k);end;begin  read(r,c);  read(n);  fillchar(v,sizeof(v),false);  for i:=1 to n do    begin      read(p[i].x,p[i].y);      v[p[i].x,p[i].y]:=true;    end;  qsort(1,n);  ans:=2;  for i:=1 to n-1 do    for j:=i+1 to n do      begin        x:=p[j].x-p[i].x;        y:=p[j].y-p[i].y;        if (p[i].x-x>0)and(p[i].y-y>0)and        (p[i].x-x<=r)and(p[i].y-y<=c) then continue;        if (p[i].x+(ans-1)*x>0)and(p[i].x+(ans-1)*x<=r)        and(p[i].y+(ans-1)*y>0)and(p[i].y+(ans-1)*y<=c) then          ans:=max(ans,dfs(p[j].x,p[j].y,x,y));      end;  if ans=2 then writeln(0)           else writeln(ans);end.