【ZJOI2017 Round1练习&BZOJ4767】D1T3 两双手（排列组合，DP）

题意：

100%的数据：|Ax|,|Ay|,|Bx|,|By| <= 500, 0 <= n,Ex,Ey <= 500

思路：听说这是一道原题

只能往右或者下走一步且有禁止点的简化版是CF559C

然而这道题并没有这么简单

以下开始转化：

转化后套用弱化版做法即可

const mo=1000000007;
var fac,exf:array[0..1000000]of int64;
    dp:array[1..1000]of int64;
    x,y:array[1..1000]of longint;
    ex,ey,sx,sy,ax,ay,bx,by:int64;
    n,n1,i,j:longint;
    u,v,eps:double;

function fabs(x:double):double;
begin
 if x<0 then exit(-x);
 exit(x);
end;

procedure swap(var x,y:longint);
var t:longint;
begin
 t:=x; x:=y; y:=t;
end;

procedure qsort(l,r:longint);
var i,j,mid1,mid2:longint;
begin
 i:=l; j:=r; mid1:=x[(l+r)>>1]; mid2:=y[(l+r)>>1];
 repeat
  while (mid1>x[i])or((mid1=x[i])and(mid2>y[i])) do inc(i);
  while (mid1<x[j])or((mid1=x[j])and(mid2<y[j])) do dec(j);
  if i<=j then
  begin
   swap(x[i],x[j]);
   swap(y[i],y[j]);
   inc(i); dec(j);
  end;
 until i>j;
 if l<j then qsort(l,j);
 if i<r then qsort(i,r);
end;

function c(x,y:longint):int64;
begin
 exit(fac[x]*exf[y] mod mo*exf[x-y] mod mo);
end;

begin
 assign(input,'hands.in'); reset(input);
 assign(output,'hands.out'); rewrite(output);
 readln(ex,ey,n1);
 readln(ax,ay,bx,by);
 fac[0]:=1; fac[1]:=1; exf[0]:=1; exf[1]:=1;
 for i:=1 to 1000000 do fac[i]:=fac[i-1]*i mod mo;
 for i:=2 to 1000000 do exf[i]:=exf[mo mod i]*(mo-mo div i) mod mo;
 for i:=1 to 1000000 do exf[i]:=exf[i-1]*exf[i] mod mo;

 eps:=1e-8;
 v:=(ex*ay-ey*ax)/(bx*ay-by*ax);
 u:=(ex*by-ey*bx)/(ax*by-ay*bx);
 if (fabs(round(v)-v)>eps)or(fabs(round(u)-u)>eps) then
 begin
  writeln(0);
  close(input);
  close(output);
  exit;
 end
  else begin inc(n); x[n]:=round(u); y[n]:=round(v); end;
 for i:=1 to n1 do
 begin
  readln(sx,sy);
  v:=(sx*ay-sy*ax)/(bx*ay-by*ax);
  u:=(sx*by-sy*bx)/(ax*by-ay*bx);
  if (fabs(round(v)-v)>eps)or(fabs(round(u)-u)>eps) then continue;
  if (u<0)or(v<0) then continue;
  if (u>x[1])or(v>y[1]) then continue;
  inc(n); x[n]:=round(u); y[n]:=round(v);
 end;
 inc(n); x[n]:=0; y[n]:=0;
 qsort(1,n);
 dp[1]:=1;
 for i:=2 to n do
 begin
  dp[i]:=c(x[i]+y[i],x[i]);
  for j:=2 to i-1 do
   if (x[i]>=x[j])and(y[i]>=y[j]) then
    dp[i]:=dp[i]-dp[j]*c(x[i]+y[i]-x[j]-y[j],x[i]-x[j]) mod mo;
  dp[i]:=(dp[i] mod mo+mo) mod mo;
 end;
 writeln(dp[n]);

 close(input);
 close(output);
end.