KM+SLACK 優化 區別於 hyc 小牛

KM+SLACK 優化 區別於  hyc 小牛  

 目前最快的KM+SLACK 傳自MLJ大牛 ，就是BFS比較難打.......

 

先初始化頂標，lx[i]是第i個節點的最大邊的權值，ly[i]都是0

 

 

調用匈牙利的BFS過程，只要注意把 a[x,i]這個條件改成 lx[x]+ly[i]=w[x,i] 即可

 

代碼如下：

 

{$A+,B-,D-,E-,F-,G+,I-,L-,N+,O-,P+,Q-,R-,S-,T-,V-,X+,Y-} {$M 65520,0,655360} {$inline on} program km; const maxlongint=1000000; var i,j,n,m,o,ans:longint; w:array[1..2000,1..2000]of longint; list,lx,ly,a,b,slack,pre:array[1..2000]of longint; vx,vy:array[1..2000]of boolean; function bfs(x1:longint):longint;inline; var i,j,t,l,r,d,x:longint; begin list[1]:=x1; l:=0; r:=1; repeat repeat inc(l); x:=list[l]; vx[x]:=true; for i:=1 to n do begin t:=lx[x]+ly[i]-w[x,i]; if (t<slack[i]) and not (vy[i]) then begin slack[i]:=t; pre[i]:=x; if slack[i]=0 then begin vy[i]:=true; if b[i]=0 then exit(i) else begin inc(r); list[r]:=b[i]; end; end; end; end; until l=r; d:=maxlongint; for i:=1 to n do if slack[i]<d then if not vy[i] then d:=slack[i]; for i:=1 to n do begin if vx[i] then dec(lx[i],d); if vy[i] then inc(ly[i],d) else dec(slack[i],d); end; for i:=1 to n do if (slack[i]=0) and not vy[i] then begin vy[i]:=true; if b[i]=0 then exit(i) else begin inc(r); list[r]:=b[i]; end; end; until l=r; bfs:=0; end; begin assign(input,'km.in'); reset(input); readln(n); for i:=1 to n do for j:=1 to n do begin read(w[i,j]); if w[i,j]>lx[i] then lx[i]:=w[i,j]; end; close(input); for i:=1 to n do begin for j:=1 to n do begin vx[j]:=false; vy[j]:=false; slack[j]:=maxlongint; end; j:=bfs(i); while j<>0 do begin o:=a[pre[j]]; a[pre[j]]:=j; b[j]:=pre[j]; j:=o; end; end; for i:=1 to n do ans:=ans+w[b[i],i]; assign(output,'km.out'); rewrite(output); writeln(ans); close(output); end.