Wormholes
Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 18166 Accepted: 6379
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.
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.
Output
Lines 1..F: For each farm, output “YES” if FJ can achieve his goal, otherwise output “NO” (do not include the quotes).
Sample Input
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
Sample Output
NO
YES
Hint
For farm 1, FJ cannot travel back in time.
For farm 2, FJ could travel back in time by the cycle 1->2->3->1, arriving back at his starting location 1 second before he leaves. He could start from anywhere on the cycle to accomplish this.
Source
USACO 2006 December Gold
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bellman_ford求負環.
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Code:
program poj_3259;
var dis:array[1..500] of longint;
s,e,t:array[1..5200] of longint;
n,m,w,i,f:longint;
function bellman_ford:boolean;
var i,j:longint; flag:boolean;
begin
for i:=1 to n do dis[i]:=maxlongint shr 1;
dis[1]:=0;
for i:=1 to n do begin
flag:=false;
for j:=1 to (m shl 1)+w do
if dis[s[j]]+t[j] flag:=true;
dis[e[j]]:=dis[s[j]]+t[j];
end;
if not flag then exit(false);
if i=n then exit(true);
end;
end;
begin
readln(f);
while f>0 do begin
dec(f);
readln(n,m,w);
for i:=1 to m do begin
readln(s[i],e[i],t[i]);
s[i+m]:=e[i];
e[i+m]:=s[i];
t[i+m]:=t[i];
end;
for i:=1 to w do begin
readln(s[(m shl 1)+i],e[(m shl 1)+i],t[(m shl 1)+i]);
t[(m shl 1)+i]:=-t[(m shl 1)+i];
end;
if bellman_ford then write(‘YES’) else write(‘NO’);
if f<>0 then writeln;
end;
end.