bellman-ford算法(判断是否存在负环)

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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;}

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