Bellman-ford算法的一个重要应用是判负环。在迭代n-1次后如果还可以进行松弛操作，说明一定存在负环。如果采用队列实现，那么当某个结点入队了n次时可以判断出存在负环，代码如下:

#include<iostream>
#include<vector>
#include<queue>
#include<cstdio>
#include<cstdlib>
#include<cstring>
using namespace std;
const int maxn = 1001;
struct Edge
{
    int from;
    int to;
    int dist;
};
struct BellmanFord
{
    int n,m;
    vector<Edge> edges;
    vector<int> G[maxn];
    bool inq[maxn]; //是否在队列中
    int d[maxn]; //s到各个点的距离
    int p[maxn]; //最短路中的上一条弧
    int cnt[maxn]; //进队次数

    void init(int n)
    {
        this->n = n;
        for(int i=0;i<n;i++) G[i].clear();
        edges.clear();
    }

    void AddEdge(int from,int to,int dist)
    {
        edges.push_back((Edge) {from,to,dist});   //输入相连通的两条边
        m = edges.size();
        G[from].push_back(m-1);
    }

    bool negetiveCycle()
    {
        queue<int> Q;
        memset(inq,0,sizeof(inq));
        memset(cnt,0,sizeof(cnt));
        for(int i=0;i<n;i++) {d[i] = 0;inq[0] = true;Q.push(i);}

        while(!Q.empty())
        {
            int u = Q.front();
            Q.pop();
            inq[u] = false;
            for(int i=0;i<G[u].size();i++)
            {
                Edge& e = edges[G[u][i]];
                if(d[e.to] > d[u] + e.dist)
                {
                    d[e.to] = d[u] + e.dist;
                    p[e.to] = G[u][i];
                    if(!inq[e.to]) {Q.push(e.to);inq[e.to] = true;if(++cnt[e.to] > n) return true;}
                }
            }
        }
        return false;
    }
};
int main()
{
    BellmanFord bell;
    int start,end,dist;
    int i,n,m;
    cout<<"please enter n vectex:"<<endl;
    //freopen("111","r",stdin);
    cin>>n;
    bell.init(n);
    cout<<"please enter m numbers vectex which two connect:"<<endl;
    cin>>m;
    for(i=0;i<m;i++)
    {
        cin>>start>>end>>dist;
        bell.AddEdge(start,end,dist);
    }
    if(bell.negetiveCycle())
    {
        cout<<"exist negative ring."<<endl;
    }
    else
    {
        cout<<"don't exist negative ring."<<endl;
    }
    return 0;
}