Ultraviolet A 11090-Going in Cycle !!

Binary + SPFA negative ring search

11090-Going in Cycle !! Time limit: 3.000 seconds

I U P C 2 0 0 6
Problem G: Going in Cycle !!
Input: standard input Output: standard output
You are given a weighted directed graphNVertices andMEdges. 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.NTest cases follow. Each one starts with two numbersNAndM.MLines follow, each has three positive numberA, B, cWhich means there is an edge from vertexAToBWith weightC.
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: Hammad Tavakoli Ghinani Alternate Solution: Cho

#include 
 
  #include 
  
   #include 
   
    #include #include 
    
     using namespace std;const double INF=1000000000.;struct Edge{    int to,next;    double w;}edge[110000];int Adj[100],Size=0,n,m;void init(){    Size=0; memset(Adj,-1,sizeof(Adj));}void Add_Edge(int u,int v,double w){    edge[Size].to=v;    edge[Size].next=Adj[u];    edge[Size].w=w;    Adj[u]=Size++;}double dist[110];int cq[110]; bool inq[110];bool spfa(double mid){    queue
     
       q;    for(int i=1;i<=n;i++)    {        q.push(i);        dist[i]=INF; cq[i]=0; inq[i]=false;    }    while(!q.empty())    {        int u=q.front(); q.pop();        inq[u]=false;        for(int i=Adj[u];~i;i=edge[i].next)        {            int v=edge[i].to;            if(dist[v]>dist[u]+edge[i].w-mid)            {                dist[v]=dist[u]+edge[i].w-mid;                if(!inq[v])                {                    inq[v]=true;                    cq[v]++;                    if(cq[v]>=n) return false;                    q.push(v);                }            }        }    }    return true;}int main(){    int T_T,cas=1;    scanf("%d",&T_T);    while(T_T--)    {        scanf("%d%d",&n,&m);        init();        double ans=INF;        double low=0,high=0,mid;        for(int i=0;i
      
       1e-3) { mid=(high+low)/2.; if(spfa(mid)==false) { ans=min(ans,mid); high=mid; } else low=mid; } printf("%.2lf\n",ans); } return 0;}