POJ---3259 Wormholes[SPFA模版]

Wormholes

 

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 20567		Accepted: 7314

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

 

 

 

 

 

 

题意：
John的农场里N块地，M条路连接两块地，W个虫洞；路是双向的，虫洞是一条单向路，会在你离开之前把你传送到目的地，
就是当你过去的时候时间会倒退Ts。我们的任务是知道会不会在从某块地出发后又回来，看到了离开之前的自己。
简化下，就是看图中有没有负权环。

 

 

code:

 

 1 #include<iostream>

 2 #include<queue>

 3 using namespace std;

 4 

 5 #define MAXN 510

 6 #define intmax 0x3fffffff

 7 

 8 int map[MAXN][MAXN];

 9 int cnt[MAXN];

10 int vst[MAXN];

11 int dis[MAXN];

12 int f,n,m,w;

13 

14 bool spfa(int st)

15 {

16     int i;

17     for(i=0;i<=n;i++)

18     {

19         vst[i]=0;

20         dis[i]=intmax;

21         cnt[i]=0;

22     }

23     int temp;

24     queue<int>Que;

25     Que.push(st);

26     dis[st]=0;

27     vst[st]=1;

28     cnt[st]=1;

29     while(!Que.empty())

30     {

31         temp=Que.front();

32         Que.pop();

33         vst[temp]=0;

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

35         {

36             if(dis[i]>dis[temp]+map[temp][i])

37             {

38                 dis[i]=dis[temp]+map[temp][i];

39                 if(!vst[i])

40                 {

41                     cnt[i]++;

42                     Que.push(i);

43                     vst[i]=1;

44                     if(cnt[i]>=n)

45                         return false;

46                 }

47             }

48         }

49     }

50     return true;

51 }

52 

53 int main()

54 {

55     int i,j;

56     int s,e,t;

57     scanf("%d",&f);

58     while(f--)

59     {

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

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

62             for(j=1;j<=n;j++)

63                 map[i][j]=intmax;

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

65         {

66             scanf("%d%d%d",&s,&e,&t);

67             if(t<map[s][e])

68             {

69                 map[s][e]=t;

70                 map[e][s]=t;

71             }

72         }

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

74         {

75             scanf("%d%d%d",&s,&e,&t);

76             map[s][e]=-t;

77         }

78         if(spfa(1))

79             printf("NO\n");

80         else

81             printf("YES\n");

82     }

83     return 0;

84 }