UVA - 11090 Going in Cycle!!

Going in Cycle!!
Time Limit: 3000MS
Memory Limit: Unknown
64bit IO Format: %lld & %llu
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Description
I I U P C 2 0 06
Problem G: Going in Cycle!!
Input: standard input Output: standard output
  
You are given a weighted directed graph with n vertices and m edges. Each cycle in the graph has a weight, which equals to sum of its edges. There are so many cycles in the graph with different weights. In this problem we want to find a cycle with the minimum mean.  
Input
The first line of input gives the number of cases, N. N test cases follow. Each one starts with two numbers n and m. m lines follow, each has three positive number a, b, c which means there is an edge from vertex a to b with weight of c.  
Output
For each test case output one line containing “Case #x: ” followed by a number that is the lowest mean cycle in graph with 2 digits after decimal place, if there is a cycle. Otherwise print “No cycle found.”.  
Constraints
-           n ≤ 50 -           a, b ≤ n -           c ≤ 10000000  
Sample Input
Output for Sample Input
2 2 1 1 2 1 2 2 1 2 2 2 1 3
Case #1: No cycle found. Case #2: 2.50
 
Problemsetter: Mohammad Tavakoli Ghinani

const int inf=0x3fffffff;
const int maxn=556;
struct Edge
{
    int from,to;double dist;
};
struct bf
{
    int n,m;
    vector<Edge> edges;
    VI G[maxn];
    bool inq[maxn];
    double d[maxn];
    int p[maxn];
    int cnt[maxn];

    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,double dist)
    {
        edges.pb((Edge){from,to,dist});
        m=edges.size();
        G[from].pb(m-1);
    }
    bool negative()
    {
        queue<int> q;

        memset(inq,0,sizeof inq);
        memset(cnt,0,sizeof cnt);
        for(int i=0;i<n;i++){d[i]=0;inq[i]=1;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;
    }
}solver;
int n,m;
bool test(double x)
{
    ///cout<<m<<endl;
    for(int i=0;i<m;i++)
        solver.edges[i].dist-=x;
    bool ret=solver.negative();
    for(int i=0;i<m;i++)
        solver.edges[i].dist+=x;
    return ret;
}
int main(){
int T;cin>>T;int kase=1;
while(T--)
{
    cin>>n>>m;
    solver.init(n);
    double ub=0;
    for(int i=0;i<m;i++)    ///while(m--)
    {
        int u,v;double w;
        scanf("%d%d%lf",&u,&v,&w);u--,v--;
        ub=max(ub,w);
        solver.addedge(u,v,w);
    }
    printf("Case #%d: ",kase++);
    if(!test(ub+1.0))puts("No cycle found.");
    else{
        double l=0,r=ub;
        while(r-l>1e-3)
        {
            double m=(l+r)/2;
            if(test(m))r=m;else l=m;
        }
        printf("%.2lf
",l);
    }
}
    return 0;
}