• bzoj 1797: [Ahoi2009]Mincut 最小割 (网络流)


    太神了直接看了hzwer的题解,有个新认识,一条路径上满流的一定是这条路径上所有边的最小值。

    type
      arr=record
        toward,next,cap,from:longint;
      end;
    const
      maxm=200000;
      maxn=10000;
    var
      edge:array[0..maxm]of arr;
      first,cur,d,p,gap:array[0..maxn]of longint;
      chose1,chose2:array[0..maxn]of boolean;
      n,m,s,t,tot,esum:longint;
     
    function min(x,y:longint):longint;
    begin
      if x<y then exit(x);
      exit(y);
    end;
     
    procedure add(j,k,l:longint);
    begin
      inc(esum);
      edge[esum].from:=j;
      edge[esum].toward:=k;
      edge[esum].next:=first[j];
      first[j]:=esum;
      edge[esum].cap:=l;
    end;
     
    procedure addedge(j,k,l:longint);
    begin
      add(j,k,l);
      add(k,j,0);
    end;
     
    function sap(x,flow:longint):longint;
    var
      now,more,i,too:longint;
    begin
      if x=t then exit(flow);
      now:=0;
      i:=cur[x];
      while i>=0 do begin
        too:=edge[i].toward;
        if (d[x]=d[too]+1) and (edge[i].cap>0) then begin
          more:=sap(too,min(flow-now,edge[i].cap));
          dec(edge[i].cap,more);
          inc(edge[i xor 1].cap,more);
          inc(now,more);
          cur[x]:=i;
          if now=flow then exit(flow);
        end;
        i:=edge[i].next;
      end;
      dec(gap[d[x]]);
      if gap[d[x]]=0 then d[s]:=n;
      inc(d[x]);
      inc(gap[d[x]]);
      cur[x]:=first[x];
      exit(now);
    end;
     
    procedure maxflow;
    var
      i:longint;
    begin
      fillchar(d,sizeof(d),0);
      fillchar(gap,sizeof(gap),0);
      gap[0]:=n;
      for i:=1 to n do cur[i]:=first[i];
      while d[s]<n do sap(s,maxlongint);
    end;
     
    procedure into;
    var
      i,j,k,l:longint;
    begin
      esum:=-1;
      fillchar(first,sizeof(first),255);
      readln(n,m,s,t);
      for i:=1 to m do begin
        readln(j,k,l);
        addedge(j,k,l);
      end;
    end;
     
    procedure work;
    var
      head,tail,x,y,i,j,too:longint;
    begin
      maxflow;
     { for i:=0 to m<<1 do
        writeln(edge[i].from,' ',edge[i].toward,' ',edge[i].next,' ',edge[i].cap); }
      head:=1;
      tail:=1;
      p[1]:=s;
      fillchar(chose1,sizeof(chose1),false);
      chose1[s]:=true;
      while head<=tail do begin
        x:=p[head];
        i:=first[x];
        while i>=0 do begin
          too:=edge[i].toward;
          if (edge[i].cap>0) and (not chose1[too]) then begin
            inc(tail);
            p[tail]:=too;
            chose1[too]:=true;
          end;
          i:=edge[i].next;
        end;
        inc(head);
      end;
      //for i:=1 to n do writeln(i,' ',chose1[i]);
      head:=1;
      tail:=1;
      p[1]:=t;
      fillchar(chose2,sizeof(chose2),false);
      chose2[t]:=true;
      while head<=tail do begin
        x:=p[head];
        i:=first[x];
        while i>=0 do begin
          too:=edge[i].toward;
          if (edge[i xor 1].cap>0) and (not chose2[too]) then begin
            inc(tail);
            p[tail]:=too;
            chose2[too]:=true;
          end;
          i:=edge[i].next;
        end;
        inc(head);
      end;
      //for i:=1 to n do writeln(i,' ',chose2[i]);
      for i:=1 to m do begin
        j:=(i-1)<<1;
        if edge[j].cap>0 then begin
          writeln(0,' ',0);
          continue;
        end;
        x:=edge[j].from;
        y:=edge[j].toward;
        if not ( (chose1[x] and chose1[y]) or (chose2[x] and chose2[y]) )then write(1) else write(0);
        write(' ');
        if (chose1[x] and chose2[y]) or (chose1[y] and chose2[x])then writeln(1) else writeln(0);
      end;
      readln;
    end;
     
    begin
      into;
      work;
    end.
    View Code
  • 相关阅读:
    Educational Codeforces Round 6
    Codeforces Round #373 (Div. 2)
    尺取法
    Codeforces Round #542 [Alex Lopashev Thanks-Round] (Div. 2)
    逆元(数论倒数)
    最大公约数gcd,最小公倍数lcm,扩展欧几里得
    hdu 6395 Sequence (分段矩阵快速幂)
    快速幂
    hdu 6432 Cyclic
    hdu 6397 charactor encoding
  • 原文地址:https://www.cnblogs.com/Macaulish/p/4358173.html
Copyright © 2020-2023  润新知