Wormholes

【题目描述】

 While exploring his many farms, Farmer Johnhas discovered a number of amazing wormholes. A wormhole is very peculiarbecause it is a one-way path that delivers you to its destination at a timethat 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 pathsand wormholes, and return to the starting field a time before his initialdeparture. Perhaps he will be able to meet himself :) .

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

【输入文件】

 Line 1: A single integer, F. F farm descriptions follow. 
Line 1 of each farm: Three space-separatedintegers respectively: N, M, and W 
Lines 2..M+1 of each farm: Three space-separated numbers (S, E, T) that describe,respectively: a bidirectional path between S and E that requires T secondsto traverse. Two fields might be connected by more than one path. 
Lines M+2..M+W+1 of each farm:Three space-separated numbers (S, E, T) that describe, respectively: A one way path from S to E thatalso moves the traveler back T seconds.

【输出文件】

Lines 1..F: For each farm, output”YES” if FJ can achieve his goal, otherwise output “NO” (donot include the quotes).

【样例输入】

 2

3 3 1

1 2 2

1 3 4

2 3 1

3 1 3

3 2 1

1 2 3

2 3 4

3 1 8

【样例输出】

 NO

YES

【样例说明】

For farm 1, FJ cannot travelback in time. 
For farm 2, FJ could travel back in time by thecycle 1->2->3->1, arriving back at his starting location 1 secondbefore he leaves. He could start from anywhere on the cycle to accomplish this.

【解题思路】

由于存在负权边，就用Bellman_ford。题目简化一下，就是看所给的图中有没有负权回路，如果有的话，输出”YES”,否则，输出”NO”。

【源代码】/pas

type rec=record
 x,y,next:longint;
 t,w:real;
end;
var
  f,n,m,s:longint;
  d:array[1..6000]of real;
  l:array[1..6000]of longint;
  a:array[1..6000]of rec;
  p:real;
function ford:boolean;
var i,k:longint;
begin
  for i:=1 to n do d[i]:=0;
  d[s]:=p;
  for k:=1 to n-1 do
  begin
    ford:=false;
    for i:=1 to n do
    if (d[a[i].x]-a[i].t)*a[i].w>d[a[i].y] then
    begin
      d[a[i].y]:=(d[a[i].x]-a[i].t)*a[i].w;
      ford:=true;
    end;
    if ford=false then exit(false);
  end;
  for i:=1 to n do
  if (d[a[i].x]-a[i].t)*a[i].w>d[a[i].y] then
  exit(true);
end;
procedure init;
var
  i:longint;
  x,y,w:Longint;
begin
  readln(n,m,s,p);
  n:=0;
  for i:=1 to m do
  begin
    with a[i*2-1] do
    begin
      read(x,y,w,t);
      next:=l[x];
      l[x]:=i*2-1;
    end;
    with a[i*2] do
    begin
      x:=a[i*2-1].y;
      y:=a[i*2-1].x;
      readln(w,t);
      next:=l[y];
      l[y]:=i*2;
    end;
  end;
  n:=m*2;
end;
begin
  init;
  if ford then
  writeln('YES')
  else
  writeln('NO');
end.