虫洞_poj3259_bellman_ford

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.