bzoj1449

竞赛图一般是把每场比赛当作一个点，然后和相应球队连边
每场比赛一赢一输对两支球队都有影响看起来不好搞
实际上我们可以先假设参加后面后面所有球队都输（每场比赛双输）
然后对每场比赛我们选择一支球队赢计算增加的收益，这样就很简单了
下面就是经典的平方流的做法

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

 var e:array[..] of node;
     q:array[..] of longint;
     s,a,b,l,w,p,cur,pre,d:array[..] of longint;
     v:array[..] of boolean;
     ans,len,i,j,x,y,n,m,t:longint;

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

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

 function spfa:boolean;
   var x,y,i,f,r:longint;
   begin
     f:=;
     r:=;
     q[]:=;
     fillchar(v,sizeof(v),false);
     for i:= to t do
       d[i]:=inf;
     d[]:=;
     while f<=r do
     begin
       x:=q[f];
       v[x]:=false;
       i:=p[x];
       while i<>- do
       begin
         y:=e[i].po;
         if e[i].flow> then
           if d[y]>d[x]+e[i].cost then
           begin
             cur[y]:=i;
             pre[y]:=x;
             d[y]:=d[x]+e[i].cost;
             if not v[y] then
             begin
               inc(r);
               q[r]:=y;
               v[y]:=true;
             end;
           end;
         i:=e[i].next;
       end;
       inc(f);
     end;
     if d[t]=inf then exit(false) else exit(true);
   end;

 function mincost:longint;
   var i,j:longint;
   begin
     mincost:=;
     while spfa do
     begin
       mincost:=mincost+d[t];
       i:=t;
       while i<> do
       begin
         j:=cur[i];
         dec(e[j].flow);
         inc(e[j xor ].flow);
         i:=pre[i];
       end;
     end;
   end;

 begin
   len:=-;
   fillchar(p,sizeof(p),);
   readln(n,m);
   for i:= to n do
     readln(w[i],l[i],a[i],b[i]);
   t:=n+m+;
   for i:= to m do
   begin
     readln(x,y);
     inc(s[x]);
     inc(s[y]);
     build(,i,,);
     build(i,x+m,,);
     build(i,y+m,,);
   end;
   for i:= to n do
   begin
     ans:=ans+a[i]*sqr(w[i])+b[i]*sqr(s[i]+l[i]);
     for j:= to s[i] do
       build(i+m,t,,(a[i]+b[i])*(*j-)+*w[i]*a[i]-*b[i]*(s[i]+l[i]));
   end;
   writeln(mincost+ans);
 end.