CSU 1659: Graph Center（SPFA）

Description

The center of a graph is the set of all vertices of minimum eccentricity, that is, the set of all vertices A where the greatest distance d(A,B) to other vertices B is minimal. Equivalently, it is the set of vertices with eccentricity equal to the graph's radius. Thus vertices in the center (central points) minimize the maximal distance from other points in the graph.
------wikipedia
Now you are given a graph, tell me the vertices which are the graph center.

Input

There are multiple test cases.
The first line will contain a positive integer T (T ≤ 300) meaning the number of test cases.
For each test case, the first line contains the number of vertices N (3 ≤ N ≤ 100) and the number of edges M (N - 1 ≤ N * (N - 1) / 2). Each of the following N lines contains two vertices x (1 ≤ x ≤ N) and y (1 ≤ y ≤ N), meaning there is an edge between x and y.

Output

The first line show contain the number of vertices which are the graph center. Then the next line should list them by increasing order, and every two adjacent number should be separated by a single space.

Sample Input

2
4 3
1 3
1 2
2 4
5 5
1 4
1 3
2 4
2 3
4 5

Sample Output

2
1 2
3
1 2 4

HINT

Source

#include 
#include 
#include 
using namespace std; 
  
const int MAXN = 105; 
const int MAXM = 100005; 
const int INF = 1<<30; 
struct EDG{ 
    int to,next; 
}edg[MAXM]; 
int eid,head[MAXN]; 
  
void init(){ 
    eid=0; 
    memset(head,-1,sizeof(head)); 
} 
void addEdg(int u,int v){ 
    edg[eid].to=v; edg[eid].next=head[u]; head[u]=eid++; 
    edg[eid].to=u; edg[eid].next=head[v]; head[v]=eid++; 
} 
int spfa(int s,int n){ 
    queueq; 
    bool inq[MAXN]={0}; 
    int d[MAXN]; 
    for(int i=1; i<=n; i++) 
        d[i]=INF; 
    d[s]=0; 
    q.push(s); 
    while(!q.empty()){ 
        int u=q.front(); q.pop(); 
        inq[s]=0; 
        for(int i=head[u]; i!=-1; i=edg[i].next){ 
            int v=edg[i].to; 
            if(d[v]>d[u]+1){ 
                d[v]=d[u]+1; 
                if(!inq[v]) 
                 q.push(v),inq[v]=1; 
            } 
        } 
    } 
    int maxt=0; 
    for(int i=1; i<=n; i++) 
        if(maxt