POJ 3259 Wormholes (最短路)

Wormholes

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 34302		Accepted: 12520

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.

 

 

 

以每个点为起点跑一次，检测是否有负环。

 1 #include <iostream>

 2 #include <cstdio>

 3 using    namespace    std;

 4 

 5 const    int    INF = 0xfffffff;

 6 const    int    SIZE = 5500;

 7 int    D[505];

 8 int    N,M,W,F;

 9 struct    Node

10 {

11     int    from,to,cost;

12 }G[SIZE];

13 

14 bool    Bellman_Ford(int);

15 bool    relax(int,int,int);

16 int    main(void)

17 {

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

19     while(F --)

20     {

21         scanf("%d%d%d",&N,&M,&W);

22         int    i = 0;

23         while(i < 2 * M)

24         {

25             scanf("%d%d%d",&G[i].from,&G[i].to,&G[i].cost);

26             i ++;

27             G[i] = G[i - 1];

28             swap(G[i].from,G[i].to);

29             i ++;

30         }

31         while(i < W + M * 2)

32         {

33             scanf("%d%d%d",&G[i].from,&G[i].to,&G[i].cost);

34             G[i].cost = -G[i].cost;

35             i ++;

36         }

37         bool    flag = true;

38         for(int i = 1;i <= N;i ++)

39             if(!Bellman_Ford(i))

40             {

41                 puts("YES");

42                 flag = false;

43                 break;

44             }

45         if(flag)

46             puts("NO");

47     }

48 

49     return    0;

50 }

51 

52 bool    Bellman_Ford(int s)

53 {

54     fill(D,D + 505,INF);

55     D[s] = 0;

56     bool    update;

57 

58     for(int i = 0;i < N - 1;i ++)

59     {

60         update = false;

61         for(int i = 0;i < M * 2 + W;i ++)

62             if(relax(G[i].from,G[i].to,G[i].cost))

63                 update = true;

64         if(!update)

65             break;

66     }

67     for(int i = 0;i < M * 2 + W;i ++)

68         if(relax(G[i].from,G[i].to,G[i].cost))

69             return    false;

70     return    true;

71 }

72 

73 bool    relax(int from,int to,int cost)

74 {

75     if(D[to] > D[from] + cost)

76     {

77         D[to] = D[from] + cost;

78         return    true;

79     }

80     return    false;

81 }

Bellman_Ford