• POJ 3259——Wormholes——————【最短路、SPFA、判负环】


    Wormholes
    Time Limit:2000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

    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..NM (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, FF farm descriptions follow. 
    Line 1 of each farm: Three space-separated integers respectively: NM, and W
    Lines 2.. M+1 of each farm: Three space-separated numbers ( SET) 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.. MW+1 of each farm: Three space-separated numbers ( SET) 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.
     
     
    题目大意:求解图中是否存在负环。题意很恶心。
     
    解题思路:SPFA判负环。当某个顶点入队n次,则说明存在负环。
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  • 原文地址:https://www.cnblogs.com/chengsheng/p/4902919.html
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