BZOJ2661 连连看 （费用流）

把所有点拆成两个，将符合条件的两个点x,y连上边，流量为1，费用为-(x+y)。

做一遍最小费用最大流，最后ans div 2即可。

 Program bzoj2661;
 const INF=;
 var last,next,p,cost,cap,q:array[..] of longint;
     flag1,flag2:array[..] of boolean;
     a,c,i,j,x,y,t,s,sum:longint;
     b:array[..] of longint;
     pd:array[..] of boolean;
     d,ps:array[..] of longint;
 function gcd(x,y:longint):longint;
 begin
   if y= then exit(x) else exit(gcd(y,x mod y));
 end;
 procedure add(x,y,f,c:longint);
 begin
   inc(sum); next[sum]:=last[x]; last[x]:=sum;
   p[sum]:=y; q[sum]:=x; cost[sum]:=c; cap[sum]:=f;
 end;
 procedure adt(x,y,f,c:longint);
 begin
   add(x,y,f,c); add(y,x,,-c);
 end;
 procedure spfa;
 var i,j,l,r:longint;
 begin
   for i:= to t  do d[i]:=INF;
   fillchar(pd,sizeof(pd),false);
   l:=; r:=; pd[s]:=true; b[]:=s; d[s]:=;
   while l<=r do
     begin
       i:=last[b[l]];
       while i<> do
         begin
           if (cap[i]>) and (d[p[i]]>d[b[l]]+cost[i]) then
             begin
               d[p[i]]:=d[b[l]]+cost[i];
               ps[p[i]]:=i;
               if not pd[p[i]] then
                 begin
                   inc(r);
                   b[r]:=p[i];
                   pd[p[i]]:=true;
                 end;
             end;
           i:=next[i];
         end;
       pd[b[l]]:=false;
       inc(l);
     end;
 end;
 procedure minCmaxF;
 var x,i,j,cc,f,min:longint;
 begin
   f:=; cc:=;
   while true do
     begin
       spfa;
       if d[t]=INF then
         begin
           writeln(f div ,' ',-cc div );
           exit;
         end;
       min:=INF;
       x:=t;
       while x<>s do
         begin
           if cap[ps[x]]<min then min:=cap[ps[x]];
           x:=q[ps[x]];
         end;
       x:=t;
       while x<>s do
         begin
           dec(cap[ps[x]],min);
           inc(cap[ps[x] xor ],min);
           x:=q[ps[x]];
         end;
       cc:=cc+min*d[t];
       f:=f+min;
     end;
 end;
 begin
   readln(a,c);
   s:=;
   for i:=a to c do
     for j:=a to c do if (i<>j) then
       begin
         x:=trunc(sqrt(abs(i*i-j*j)));
         if x*x<>abs(i*i-j*j) then continue;
         if gcd(i,j)<> then continue;
         add(i,j+c,,-(i+j));
       end;
  for i:=a to c do add(,i,,);
  for i:=a to c do add(i+c,c*+,,);
  s:=; t:=*c+;
  minCmaxF;
 end.