二分+SPFA找負環

11090 - Going in Cycle!!

Time limit: 3.000 seconds

I 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 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

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;i1e-3)
        {
            mid=(high+low)/2.;
            if(spfa(mid)==false)
            {
                ans=min(ans,mid); high=mid;
            }
            else low=mid;
        }
        printf("%.2lf\n",ans);
    }
    return 0;
}