poj 3259 Wormholes（虫洞）

题目

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.
就是让找有没有负权回路，因为只要有负权回路，FJ就一定可以在0时刻之前回到起点

题解

判断有没有负权回路，用Bellman_ford
过程：做n次循环每一条边，判断是否可以用它更新到某个点的最短路
然后判断是不是存在一条边每加一次都可以让最短路变小，若存在，则存在负权回路

时间复杂度O(NE)。E为边数

代码

var  n,m,w,f,i,j,k,l:longint;  e:array[1..3,1..6000]of longint;  d:array[1..500]of longint;function bellman_ford(n,m:longint):boolean;var  i,j,f:longint;begin  fillchar(d,sizeof(d),$7f);  d[1]:=0;  for i:=1 to n do    begin      f:=1;      for j:=1 to m do        if d[e[2,j]]+e[3,j]<d[e[1,j]] then begin f:=0;                                                 d[e[1,j]]:=d[e[2,j]]+e[3,j];                                           end;      if f=1 then break;    end;  for i:=1 to m do    if d[e[2,i]]+e[3,i]<d[e[1,i]] then exit(false);  exit(true);end;begin  readln(f);  for l:=1 to f do    begin      k:=0;      readln(n,m,w);      for i:=1 to m do        begin          inc(k);          readln(e[1,k],e[2,k],e[3,k]);          inc(k);          e[1,k]:=e[2,k-1];e[2,k]:=e[1,k-1];e[3,k]:=e[3,k-1];        end;      for i:=1 to w do        begin          inc(k);          readln(e[1,k],e[2,k],j);          e[3,k]:=-j;        end;      if bellman_ford(n,k) then writeln('NO') else      writeln('YES');    end;end.