poj Wormholes （最短路 spfa Bellman-Ford 算法 邻接表实现）

Wormholes

Time Limit:                                                        2000MS

Memory Limit: 65536KB

64bit IO Format:                            %I64d & %I64u

Submit                                        Status                             

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 toF (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.

#include
#include
#include
#include
using namespace std;
const int INF=505;
int inq[INF];
int cnt[INF];
int low[INF];
int head[INF];
int n,m,c;
struct Edge
{
    int next_edge,v,t;
} edge[INF*INF];

bool spfa(int arc)
{
    queueq;
    q.push(arc);
    cnt[arc]++;
    memset(inq,0,sizeof(inq));
    memset(cnt,0,sizeof(cnt));
    memset(low,0x1f,sizeof(low));
    low[arc]=0;
    while(!q.empty())
    {
        int  x=q.front();
        q.pop();
        inq[x]=0;
        for(int e=head[x]; e!=-1; e=edge[e].next_edge)
        {
            if(low[edge[e].v]>low[x]+edge[e].t)
            {
                low[edge[e].v]=edge[e].t+low[x];

                if(!inq[edge[e].v])
                {
                    inq[edge[e].v]=1;
                    q.push(edge[e].v);
                    if(++cnt[edge[e].v]>n)
                        return 1;
                }
            }
        }

    }
    return 0;
}

int main()
{
    int t;
    int x,y,z;
    cin>>t;
    while(t--)
    {
        memset(head,-1,sizeof(head));
        cin>>n>>m>>c;
        for(int i=0; i>x>>y>>z;
            edge[i].next_edge=head[x];
            edge[i].v=y;
            edge[i].t=z;
            head[x]=i;

            edge[i+m].next_edge=head[y];
            edge[i+m].v=x;
            edge[i+m].t=z;
            head[y]=i+m;
        }

        for(int i=0; i>x>>y>>z;
            edge[i+2*m].next_edge=head[x];
            edge[i+2*m].v=y;
            edge[i+2*m].t=-z;
            head[x]=i+2*m;
        }
        printf("%s\n",spfa(1)?"YES":"NO");
    }
    return 0;
}