Poj 3259 Wormholes判断负权回路（spfa模板题）

While exploring his many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it is a one-way path that delivers you to its destination at a time that is BEFORE you entered the wormhole! Each of FJ's farms comprises N (1 ≤ N ≤ 500) fields conveniently numbered 1..N, M (1 ≤ M ≤ 2500) paths, and W (1 ≤ W ≤ 200) wormholes.

As FJ is an avid time-traveling fan, he wants to do the following: start at some field, travel through some paths and wormholes, and return to the starting field a time before his initial departure. Perhaps he will be able to meet himself :) .

To help FJ find out whether this is possible or not, he will supply you with complete maps to F (1 ≤ F ≤ 5) of his farms. No paths will take longer than 10,000 seconds to travel and no wormhole can bring FJ back in time by more than 10,000 seconds.

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    朴素的SPFA，判断有无负权回路即可。注意普通道路是无向的。完美的打一遍模板，AK。

var  t:boolean;  f,i,n:longint;  way:array[1..500,1..500] of longint;function min(a,b:longint):longint;begin  if a<=b then exit(a) else exit(b);end;procedure Readin;var  i,j,si,ei,ti,m,w:longint;begin  read(n,m,w);  for i:=1 to n do for j:=1 to n do way[i,j]:=maxlongint;  for i:=1 to m+w do beginread(si,ei,ti);if i>m then ti:=-ti;way[si,ei]:=min(ti,way[si,ei]);if i<=m then way[ei,si]:=way[si,ei];  end;end;procedure SPFA;var  time,dis:array [1..500] of longint;  used:array[1..500] of boolean;  que:array[1..1000000] of longint;  head,tail,i,x:longint;begin  t:=false;  dis[1]:=0;for i:=2 to n do dis[i]:=maxlongint;  fillchar(used,sizeof(used),false);  fillchar(time,sizeof(time),0);  head:=1;tail:=1;que[1]:=1;time[1]:=1;  while head<=tail do begin    x:=que[head];for i:=1 to n do  if (way[x,i]<>maxlongint) and (dis[i]>dis[x]+way[x,i]) then begin    dis[i]:=dis[x]+way[x,i];if not used[i] then begin  if time[i]+1>n then begin    t:=true;break;  end;  inc(time[i]);  inc(tail);  que[tail]:=i;end;  end;used[head]:=false;inc(head);  end;end;procedure Print;begin  if t then writeln('YES')  else writeln('NO');end;begin  read(f);  for i:=1 to f do beginReadin;SPFA;Print;  end;end.