poj3259 Bellman_Ford算法

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 34465		Accepted: 12585

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.

 

 

!@$!#%^#$^@#%$^&*^!@$#@#$%^&**(&^%!^&*&!

 1 #include<stdio.h>

 2 #include<string.h>

 3 

 4 const int INF=99999999;

 5 

 6 struct Node{

 7     int u;

 8     int v;

 9     int o;

10 };

11 

12 Node  edge[10250];

13 int dis[550];

14 int n,m,w,nn;

15 

16 void relax(int u,int v,int o)

17 {

18     if(dis[v]>dis[u]+o)

19         dis[v]=dis[u]+o;

20 }

21 

22 bool Bellan_Ford()

23 {

24     int i,j,k;

25     for(i=1;i<=n;i++)

26         dis[i]=INF;

27 

28     for(i=1;i<=n-1;i++)

29     {

30         for(j=1;j<=nn;j++)

31             relax(edge[j].u,edge[j].v,edge[j].o);

32     }

33 

34     for(i=1;i<=nn;i++)

35     {

36         if(dis[edge[i].v]>dis[edge[i].u]+edge[i].o)

37         {

38             return false;

39         }

40     }

41     return true;

42 }

43 

44 int main()

45 {

46     int F,i,j,k,flg;

47     int s,e,t;

48     scanf("%d",&F);

49     while(F--)

50     {

51         flg=0;nn=1;

52         scanf("%d %d %d",&n,&m,&w);

53         for(i=1;i<=m;i++)

54         {

55             scanf("%d %d %d",&edge[nn].u,&edge[nn].v,&edge[nn].o);

56             nn++;

57             edge[nn].u=edge[nn-1].v,edge[nn].v=edge[nn-1].u,edge[nn].o=edge[nn-1].o;

58             nn++;

59         }

60         for(i=1;i<=w;i++)

61         {

62             scanf("%d %d %d",&edge[nn].u,&edge[nn].v,&edge[nn].o);

63             edge[nn].o=-edge[nn].o;

64             nn++;

65         }

66         nn--;

67         if(Bellan_Ford()==false)

68         {

69             flg=1;

70         }

71         if(flg==1)

72             printf("YES\n");

73         else

74             printf("NO\n");

75     }

76     return 0;

77 }

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