CSU1659: Graph Center (Shortest Path)

CSU1659: Graph Center (Shortest Path)
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 in 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 shocould list them by increasing order, and every two adjacent number shocould be separated by a single space.

Sample Input

24 31 31 22 45 51 41 32 42 34 5

Sample Output

21 231 2 4

HINT

 

Source
Question: give n vertices and m edges, find the distance between each vertex and other vertices, obtain the maximum value, and then select a minimum value for all the maximum distances, finally, what points under this value are output to meet the conditions?

Train of Thought: Find out all the answers in the nth Shortest Path

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              using namespace std;#define lson 2*i#define rson 2*i+1#define LS l,mid,lson#define RS mid+1,r,rson#define UP(i,x,y) for(i=x;i<=y;i++)#define DOWN(i,x,y) for(i=x;i>=y;i--)#define MEM(a,x) memset(a,x,sizeof(a))#define W(a) while(a)#define gcd(a,b) __gcd(a,b)#define LL long long#define N 200005#define INF 0x3f3f3f3f#define EXP 1e-8#define lowbit(x) (x&-x)const int mod = 1e9+7;const int L = 10005;struct Edges{ int x,y,w,next;} e[L<<2];int head[L],n,m;int dis[L];int vis[L];int cnt[L],hash[L],ss[L];int s[L];void init(){ memset(e,-1,sizeof(e)); memset(head,-1,sizeof(head));}void AddEdge(int x,int y,int w,int k){ e[k].x = x,e[k].y = y,e[k].w = w,e[k].next = head[x],head[x] = k;}int relax(int u,int v,int c){ if(dis[v]>dis[u]+c) { dis[v] = dis[u]+c; return 1; } return 0;}int SPFA(int src){ int i; memset(vis,0,sizeof(vis)); for(int i = 0; i<=n; i++) dis[i] = INF; dis[src] = 0; queue
              
                Q; Q.push(src); vis[src] = 1; while(!Q.empty()) { int u,v; u = Q.front(); Q.pop(); vis[u] = 0; for(i = head[u]; i!=-1; i=e[i].next) { v = e[i].y; if(relax(u,v,e[i].w)==1 && !vis[v]) { Q.push(v); vis[v] = 1; } } } int maxn = -1; for(i = 1; i<=n; i++) maxn = max(maxn,dis[i]); return maxn;}int ans[L],tot,p[N];int main(){ int t,u,v,i,j,k; scanf("%d",&t); while(t--) { scanf("%d%d",&n,&m); init(); for(i = 0; i<2*m; i+=2) { scanf("%d%d",&u,&v); AddEdge(u,v,1,i); AddEdge(v,u,1,i+1); } int minn = INF; for(i = 1; i<=n; i++) { p[i] = SPFA(i); minn = min(p[i],minn); } tot = 0; for(i = 1; i<=n; i++) { if(p[i]==minn) ans[tot++] = i; } printf("%d\n",tot); for(i = 0; i