CSU1659: Graph Center(最短路)

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

24 31 31 22 45 51 41 32 42 34 5

Sample Output

21 231 2 4

HINT

Source

题意：

给出n个点，m条边，求每个点到其他点的距离，取最大的，然后在这所有最大的距离中选一个最小的值，最后输出这个值下有哪些点符合条件

思路：

n次最短路找出所有答案

#include <iostream>#include <stdio.h>#include <string.h>#include <string>#include <stack>#include <queue>#include <map>#include <set>#include <vector>#include <math.h>#include <bitset>#include <list>#include <algorithm>#include <climits>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<int> 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<tot; i++)        {            if(i)                printf(" ");            printf("%d",ans[i]);        }        printf("\n");    }    return 0;}