bzoj3996

把这个式子弄清楚就知道这是最小割了

相当于，选某个点i有收入ai,i，会损失ci,

如果i,j都被选则有额外收入ai,j+aj,i

明显，对每个点i，连(s,i,∑ai,j) (i,t,ci)

对每对i,j连边(i,j,ai,j)，没了

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

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

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

 procedure build(x,y,f:longint);
   begin
     add(x,y,f);
     add(y,x,);
   end;

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

 function sap:longint;
   var u,i,j,tmp,neck,q:longint;
   begin
     numh[]:=t+;
     for i:= to t do
       cur[i]:=p[i];
     u:=; sap:=; neck:=inf;
     while h[]<t+ do
     begin
       d[u]:=neck;
       i:=cur[u];
       while i<>- do
       begin
         j:=e[i].po;
         if (e[i].flow>) and (h[u]=h[j]+) then
         begin
           neck:=min(neck,e[i].flow);
           pre[j]:=u;
           cur[u]:=i;
           u:=j;
           if u=t then
           begin
             sap:=sap+neck;
             while u<> do
             begin
               u:=pre[u];
               j:=cur[u];
               dec(e[j].flow,neck);
               inc(e[j xor ].flow,neck);
             end;
             neck:=inf;
           end;
           break;
         end;
         i:=e[i].next;
       end;
       if i=- then
       begin
         dec(numh[h[u]]);
         if numh[h[u]]= then break;
         q:=-;
         tmp:=t;
         i:=p[u];
         while i<>- do
         begin
           j:=e[i].po;
           if e[i].flow> then
             if tmp>h[j] then
             begin
               q:=i;
               tmp:=h[j];
             end;
           i:=e[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
     s:=;
     for j:= to n do
     begin
       read(x);
       s:=s+x;
       build(i,j,x);
     end;
     build(,i,s);
     ans:=ans+s;
   end;
   for i:= to n do
   begin
     read(x);
     build(i,t,x);
   end;
   writeln(ans-sap);
 end.