UVA 10048 A - Wormholes

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.

农场出现了单向的虫洞农场主john能否通过虫洞看到原来的自己。
用bellman ford判断是否有负环。

#include<stdio.h>#include<string.h>#include<iostream>#include<algorithm>using namespace std;int N,M,W,Count;int s[5400],e[5400],t[5400];int d[5400];void Bellman_Ford(){    d[1]=0;    int i=0;    bool flag=false;     for(int i=0;i<=N;i++)     {         flag=false;         for(int j=0;j<Count;j++)         {             if(d[e[j]]>d[s[j]]+t[j])             {                 d[e[j]]=d[s[j]]+t[j];                 flag=true;             }         }         if(!flag)             break;     }     for(int j=0;j<Count;j++)     {         if(d[e[j]]>d[s[j]]+t[j])         {             cout<<"YES"<<endl;             return;         }     }     cout<<"NO"<<endl;}int main(){    int F;    cin>>F;    while(F--)    {        Count=0;        cin>>N>>M>>W;        for(int i=0;i<2*(M+W);i++)        {            s[i]=99999;            e[i]=99999;            t[i]=99999;            d[i]=99999;        }        for(int i=0;i<M;i++)        {            cin>>s[Count]>>e[Count]>>t[Count];            Count++;            s[Count]=e[Count-1];            e[Count]=s[Count-1];            t[Count]=t[Count-1];            Count++;        }        for(int i=0;i<W;i++)        {            cin>>s[Count]>>e[Count]>>t[Count];            t[Count]=-t[Count];            Count++;        }        Bellman_Ford();    }    return 0;}