PKU2387 Til the Cows Come Home

     一丝不挂的最短路，我只是想试一下传说中的spfa优化——slf&lll。自我感觉是这么一个意思，但是怎么没发现这么优化之后快多少呢？有哪位牛懂这两个优化的能否赐教一下？

    当然优化的动机是很明显的，用尽量短的距离去修改更容易更早地获得最短路径。

    另外，这也是我第一次用前向星，感觉比邻接表方便些… 

 program Til_the_Cows_Come_Home; type rec=record x,y,w:longint; end; const inf=200000000; var edge:array[1..4000] of rec; dist,f:array[1..1000] of longint; que:array[-5000000..5000000] of longint; vis:array[1..1000] of boolean; head,tail,v,u,dd,dsum,i,a,b,c,n,m:longint; procedure qsort(s,t:longint); var l,r,m:longint; tt:rec; begin l:=s; r:=t; m:=edge[random(t-s+1)+s].x; while l<=r do begin while edge[l].x<m do inc(l); while edge[r].x>m do dec(r); if l<=r then begin tt:=edge[l]; edge[l]:=edge[r]; edge[r]:=tt; inc(l); dec(r); end; end; if s<r then qsort(s,r); if l<t then qsort(l,t); end; begin assign(input,'a.in'); reset(input); assign(output,'a.out'); rewrite(output); readln(m,n); for i:=1 to m do begin readln(a,b,c); edge[2*i-1].x:=a; edge[2*i-1].y:=b; edge[2*i-1].w:=c; edge[2*i].x:=b; edge[2*i].y:=a; edge[2*i].w:=c; end; randomize; qsort(1,2*m); for i:=2*m downto 1 do f[edge[i].x]:=i; f[n+1]:=2*m+1; for i:=n downto 1 do if f[i]=0 then f[i]:=f[i+1]; fillchar(vis,sizeof(vis),0); filldword(dist,sizeof(dist) shr 2,inf); head:=1; tail:=1; vis[1]:=true; dist[1]:=0; que[1]:=1; dsum:=0; repeat dd:=dsum div (tail-head+1); while dist[que[head]]>dd do begin inc(tail); que[tail]:=que[head]; inc(head); end; v:=que[head]; vis[v]:=false; dec(dsum,dist[v]); inc(head); for i:=f[v] to f[v+1]-1 do if dist[v]+edge[i].w<dist[edge[i].y] then begin u:=edge[i].y; dist[u]:=dist[v]+edge[i].w; if not vis[u] then begin vis[u]:=true; inc(dsum,dist[u]); if (head>tail) or (dist[u]>dist[que[head]]) then begin inc(tail); que[tail]:=u; end else begin dec(head); que[head]:=u; end; end; end; until head>tail; writeln(dist[n]); close(input); close(output); end.