题意/Description：
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.
(
呵呵，看不懂？找谷歌翻译。)
读入/Input：
Line 1: A single integer, F. F farm
descriptions follow.
Line 1 of each farm: Three space-separated integers 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 seconds
to 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 that also moves the traveler back T seconds.
（怎么还是英文！）
输出/Output：
Lines
1..F: For each farm, output "YES" if FJ can achieve his goal, otherwise output "NO" (do not include the quotes).
（终于看懂了一些。）
题解/solution：
这题的算法太明显了，题目中都讲到有负权了，当然用bellman啦。只不过多循环了几遍，小意思。
代码/Code：
const
maxE=100000;
maxV=2000000;
type
arr=record
x,y,w,next:longint;
end;
var
n,m,wt,t:longint;
a:array [0..maxV] of arr;
d:array [0..maxE] of longint;
function bellman:boolean;
var
i,j:longint;
begin
for i:=1 to n do
for j:=1 to m do
with a[j] do
if d[x]+w<d[y] then
d[y]:=d[x]+w;
for i:=1 to m do
with a[i] do
if d[x]+w<d[y] then exit(true);
bellman:=false;
end;
procedure init;
var
i,j,o,p:longint;
boo:boolean;
begin
readln(t);
for i:=1 to t do
begin
fillchar(a,sizeof(a),0);
fillchar(d,sizeof(d),$7f div 3);
readln(n,m,wt);
for j:=1 to m do
begin
o:=j*2; p:=j*2-1;
with a[p] do
begin
read(x,y,w);
a[o].x:=y; a[o].y:=x; a[o].w:=w;
end;
end;
for j:=1 to wt do
with a[j+m*2] do
begin
read(x,y,w);
w:=-w;
end;
m:=m*2+wt; d[1]:=0;
boo:=bellman;
if boo then writeln('YES')
else writeln('NO');
end;
end;
begin
init;
end.