Wormholes-poj 3259 (Bellman-ford algorithm)

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 34934		Accepted: 12752

Description

While exploring he many farms, Farmer John has discovered a number of amazing wormholes. A wormhole is very peculiar because it's a one-way path that delivers the IT destination at a time that's before you entered the wormhole! Each of the 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 a avid time-traveling fan, he wants to does the following:start at some field, travel through some paths and worm Holes, and return to the starting field a time before his initial departure. Perhaps he'll be able to meet himself:).

To help FJ find out whether this is possible or not, he'll supply you with complete maps to F (1≤ f ≤ 5) of his farms. No paths'll take longer than seconds to travel and no wormhole can bring FJ back in time by more than-seco Nds.

Input

Line 1: A single integer, F. FFarm 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 Sand EThat requires TSeconds to traverse. The might is 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-path from STo EThat also moves the traveler back TSeconds.

Output

Lines 1.. F: For each farm, output "YES" if FJ can achieve his goal, otherwise output "NO" (does not include the quotes).

Sample Input

23 3 11 2 21 3 42 3 13 1 33 2 11 2 32 3 43 1 8

Sample Output

NOYES

Hint

For farm 1, FJ cannot travel back in time.
For Farm 2, FJ could travel back on time by the cycle 1->2->3->1, arriving back at he starting location 1 second Before he leaves. He could start from anywhere on the cycle to accomplish this. The problem uses the Bellman-ford algorithm

1#include <iostream>2 using namespacestd;3 structFarm {4     intS;5     intE;6     intT;7} f[5500];8 intMain () {9     intnum;Ten     intN, M, W; OneCIN >>num; A     intf[ -]; -      for(inti =0; i < num; i++) { -CIN >> N >> M >>W; the          for(intj =0; J < N; J + +) { -F[J] =20000; -         } -f[0] =0; +          for(intj =0; J < M; J + +) { -             intA, B, C; +Cin >> a >> b >>C; Af[2*J]. S =A; atf[2*J]. Cb; -f[2*J]. T =C; -f[2*j+1]. S =b; -f[2*j+1]. E =A; -f[2*j+1]. T =C; -  in         } -          for(intj =2M J <2*m + W; J + +) { to             intA, B, C; +Cin >> a >> b >>C; -F[J]. S =A; theF[J]. E =b; *F[J]. T =0-C; $         }Panax Notoginseng          for(intj =0; J < N-1; J + +) { -              for(intK =0; K <2*m + W; k++) { the                 if(F[f[k]. E] > F[f[k]. S] +F[k]. T) { +F[F[K]. E] = F[f[k]. S] +F[k]. T; A                 } the             } +         } -         intFlag =0; $          for(intK =0; K <2*m + W; k++) { $             if(F[f[k]. E] >f[f[k]. S] +F[k]. T) { -F[F[K]. E] = F[f[k]. S] +F[k]. T; -flag=1; the                  Break; -             }Wuyi         } the         if(flag) { -cout<<"YES"<<Endl; Wu}Else{ -cout<<"NO"<<Endl; About         } $     } -     return 0; -}

Wormholes-poj 3259 (Bellman-ford algorithm)