bzoj3275

容易想到是最小割（最大权独立集）
然后每个数拆成两个点，不能同时选的之间连边跑最小割，
最后答案=总数-mincut/2 因为这样建图将流量变成了原来的两倍

当然这道题更好的建图方法是分成奇数和偶数两个集合
不难发现，奇数和奇数，偶数和偶数之间显然不能同时满足条件的
所以这样直接ans=总点数-mincut

 const inf=;
 type node=record
        point,flow,next:longint;
      end;

 var edge:array[..] of node;
     a,p,cur,pre,d,numh,h:array[..] of longint;
     s,i,j,n,t,len:longint;
     x:double;

 function gcd(a,b:longint):longint;
   begin
     if b= then exit(a)
     else exit(gcd(b,a mod b));
   end;

 function min(a,b:longint):longint;
   begin
     if a>b then exit(b) else exit(a);
   end;

 procedure add(x,y,f:longint);
   begin
     inc(len);
     edge[len].point:=y;
     edge[len].flow:=f;
     edge[len].next:=p[x];
     p[x]:=len;
   end;

 function sap:longint;
   var q,u,i,j,neck,tmp:longint;
   begin
     sap:=;
     u:=;
     numh[]:=t+;
     for i:= to t do
       cur[i]:=p[i];
     neck:=inf;
     while h[]<t+ do
     begin
       d[u]:=neck;
       i:=cur[u];
       while i<>- do
       begin
         j:=edge[i].point;
         if (edge[i].flow>) and (h[u]=h[j]+) then
         begin
           pre[j]:=u;
           cur[u]:=i;
           neck:=min(neck,edge[i].flow);
           u:=j;
           if u=t then
           begin
             sap:=sap+neck;
             while u<> do
             begin
               u:=pre[u];
               j:=cur[u];
               dec(edge[j].flow,neck);
               inc(edge[j xor ].flow,neck);
             end;
             neck:=inf;
           end;
           break;
         end;
         i:=edge[i].next;
       end;
       if i=- then
       begin
         dec(numh[h[u]]);
         if numh[h[u]]= then exit;
         tmp:=t;
         i:=p[u];
         q:=-;
         while i<>- do
         begin
           j:=edge[i].point;
           if edge[i].flow> then
             if h[j]<tmp then
             begin
               tmp:=h[j];
               q:=i;
             end;
           i:=edge[i].next;
         end;
         h[u]:=tmp+;
         inc(numh[h[u]]);
         cur[u]:=q;
         if u<> then
         begin
           u:=pre[u];
           neck:=d[u];
         end;
       end;
     end;
   end;

 begin
   len:=-;
   fillchar(p,sizeof(p),);
   readln(n);
   t:=*n+;
   for i:= to n do
   begin
     read(a[i]);
     add(,i,a[i]);
     add(i,,);
     add(i+n,t,a[i]);
     add(t,i+n,);
     s:=s+a[i];
   end;
   for i:= to n- do
     for j:=i+ to n do
     begin
       x:=sqrt(sqr(a[i])+sqr(a[j]));
       if (gcd(a[i],a[j])=) and (x=trunc(x)) then
       begin
         add(i,j+n,inf);
         add(j+n,i,);
         add(j,i+n,inf);
         add(i+n,j,);
       end;
     end;
   writeln(s-sap div );
 end.