SGU194 无源汇上下界可行流 上下界网络流 pascal

@2004年集训队论文 

一种简易的方法求解流量有上下界的网络中网络流问题

的问题1.1

一开始建图建反了T T

论文里很详细..不多说了=w=

code:

var  tmp,ans,tot,n,m,i,j,s,t,x,y,z,z1,z2,min,lo,p,flow:longint;  su,path,dis,anti,next,his,pre,di,vh,a,b,c,flo:array[0..100009] of longint;    flag:boolean;begin  tot:=0;ans:=0;tmp:=0;  fillchar(su,sizeof(su),0);  fillchar(b,sizeof(b),0);  read(n,m);  for i:=1 to m do    begin      read(x,y,z1,z2);      flo[i]:=z1;      su[x]:=su[x]-z1;      su[y]:=su[y]+z1;      tot:=tot+1;a[tot]:=y;c[tot]:=z2-z1;next[tot]:=b[x];b[x]:=tot;anti[tot]:=tot+1;      tot:=tot+1;a[tot]:=x;c[tot]:=0;next[tot]:=b[y];b[y]:=tot;anti[tot]:=tot-1;    end;  s:=n+1;t:=n+2;  for i:=1 to n do    if su[i]>0 then      begin        tot:=tot+1;a[tot]:=i;c[tot]:=su[i];next[tot]:=b[s];b[s]:=tot;anti[tot]:=tot+1;        tot:=tot+1;a[tot]:=s;c[tot]:=0;next[tot]:=b[i];b[i]:=tot;anti[tot]:=tot-1;        tmp:=tmp+su[i];      end    else      begin        tot:=tot+1;a[tot]:=t;c[tot]:=-su[i];next[tot]:=b[i];b[i]:=tot;anti[tot]:=tot+1;        tot:=tot+1;a[tot]:=i;c[tot]:=0;next[tot]:=b[t];b[t]:=tot;anti[tot]:=tot-1;      end;  for i:=1 to n+2 do    di[i]:=b[i];  vh[0]:=n+2;  fillchar(dis,sizeof(dis),0);  s:=n+1;t:=n+2;  i:=s;flow:=maxlongint;  while dis[s]<=(n+1) do    begin      flag:=false;his[i]:=flow;p:=di[i];      while p<>0 do        begin          j:=a[p];z:=c[p];          if ((dis[j]+1)=dis[i]) and (z>0) then            begin              if z<flow then flow:=z;              path[j]:=p;pre[j]:=i;              di[i]:=p;flag:=true;i:=j;              if i=t then                begin                  ans:=ans+flow;                  while i<>s do                    begin                      p:=path[i];c[p]:=c[p]-flow;                      c[anti[p]]:=c[anti[p]]+flow;                      i:=pre[i];                    end;                  flow:=maxlongint;                end;              break;            end;          p:=next[p];        end;      if flag then continue;      min:=t;p:=b[i];      while p<>0 do        begin          j:=a[p];          if (dis[j]<min) and (c[p]>0) then            begin              min:=dis[j];lo:=p;            end;          p:=next[p];        end;      di[i]:=lo;      vh[dis[i]]:=vh[dis[i]]-1;      if vh[dis[i]]=0 then break;      dis[i]:=min+1;inc(vh[min+1]);      if i<>s then begin i:=pre[i];flow:=his[i];end;    end;  if ans<tmp then writeln('NO') else begin  writeln('YES');  for i:=1 to m do    begin      flo[i]:=flo[i]+c[i*2];      writeln(flo[i]);    end;end;end.