Poj 3259 wormholes

1.Link:

http://poj.org/problem?id=3259

2.Content:

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 32078		Accepted: 11651

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 in the cycle to accomplish this.

Source

Usaco 2006 December Gold

3.Method:

The template title of the Bellman algorithm is mainly used to find out if the negative ring exists and has nothing to do with the initial value of dis.

4.Code:

1#include <iostream>2#include <cstring>3 4 using namespacestd;5 6 //const int max_d = 0;7 8 structRoad9 {Ten     ints; One     inte; A     intT; - }; -  the intMain () - { -     //freopen ("D://input.txt "," R ", stdin); -  +     inti,j; -  +     intff; ACIN >>ff; at  -     intn,m,w; -      while(ff--) -     { -CIN >> n >> M >>W; -  inRoad *arr_road =NewRoad[m *2+W]; -  to         ints,e,t; +          for(i =0; I < m; ++i) -         { theCIN >> s >> e >>T; *              $Arr_road[i *2].S =s;Panax NotoginsengArr_road[i *2].E =e; -Arr_road[i *2].t =T; the  +Arr_road[i *2+1].S =e; AArr_road[i *2+1].E =s; theArr_road[i *2+1].t =T; +         } -          for(i =0; i < W; ++i) $         { $CIN >> s >> e >>T; -  -Arr_road[m *2+ I].S =s; theArr_road[m *2+ I].E =e; -Arr_road[m *2+ i].t =-T;Wuyi         } the  -         //For (i = 0; i < m * 2 + W; ++i) Wu         //{ -         //cout << i << "<< arr_road[i].s <<" "<< arr_road[i].e <<" "<< arr_ road[i].t << Endl; About         //} $         //cout << Endl; -  -         int*arr_d =New int[n]; -         //For (i = 0; i < n; ++i) arr_d[i] = max_d; Amemset (Arr_d,0,sizeof(int) *n); +  the         //Bellman-ford -         BOOLFlag; $          for(i =0; I < n-1; ++i) the         { theFlag =false; the              for(j =0; J <2* m + W; ++j) the             { -                 if(ARR_D[ARR_ROAD[J].E] > ARR_D[ARR_ROAD[J].S] +arr_road[j].t) in                 { theARR_D[ARR_ROAD[J].E] = Arr_d[arr_road[j].s] +arr_road[j].t; theFlag =true; About                 } the             } the             if(!flag) Break; the         } +  -          for(j =0; J <2* m + W; ++j) the         {Bayi             if(ARR_D[ARR_ROAD[J].E] > Arr_d[arr_road[j].s] + arr_road[j].t) Break; the         } the  -         if(J <2* m + W) cout <<"YES"<<Endl; -         Elsecout <<"NO"<<Endl; the  the Delete [] arr_d; the Delete [] arr_road; the     } -  the     return 0; the}

5.Reference:

Poj 3259 wormholes