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


题意:
给出n个点。m条边,求每一个点到其它点的距离,取最大的,然后在这全部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次最短路找出全部答案

#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;
}

CSU1659: Graph Center(最短路)的更多相关文章

  1. csu - 1659 Graph Center(最短路)

    http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1659 题意是找一个图的中心,图的中心定义是某一个点到其他点的最大距离最小,如果有多个排序输出. 注 ...

  2. CSU 1659: Graph Center(SPFA)

    1659: Graph Center Time Limit: 1 Sec  Memory Limit: 128 MB Submit: 63  Solved: 25 [id=1659"> ...

  3. HDU 5876 Sparse Graph BFS 最短路

    Sparse Graph Problem Description   In graph theory, the complement of a graph G is a graph H on the ...

  4. Codeforces 715B & 716D Complete The Graph 【最短路】 (Codeforces Round #372 (Div. 2))

    B. Complete The Graph time limit per test 4 seconds memory limit per test 256 megabytes input standa ...

  5. HDU 4725 The Shortest Path in Nya Graph (最短路)

    The Shortest Path in Nya Graph Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K ...

  6. HDU4725:The Shortest Path in Nya Graph(最短路)

    The Shortest Path in Nya Graph Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K ...

  7. hdu 4725 The Shortest Path in Nya Graph (最短路+建图)

    The Shortest Path in Nya Graph Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K ...

  8. HDU 4725 The Shortest Path in Nya Graph(最短路拆点)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4725 题意:n个点,某个点属于某一层.共有n层.第i层的点到第i+1层的点和到第i-1层的点的代价均是 ...

  9. hdu4725 The Shortest Path in Nya Graph【最短路+建图】

    转载请注明出处,谢谢:http://www.cnblogs.com/KirisameMarisa/p/4297574.html      ---by 墨染之樱花 题目链接:http://acm.hdu ...

随机推荐

  1. LINQ Enumerable 续

    筛选序列 Enumerable.Distinct 对于复杂的对象列表,运行时引擎如何才能通过比较确定两个对象是否重复?对于复杂对象,必须提供一个比较器,即实现IEqualityComparer(Of ...

  2. javascript基础之for循环

    1.数组定义声名 var arry = [1,2,3,4,5]   //相当与var arry = Array(1,2,3,4,5) 2.数据的增删改查 var arry = [1,2,3,4,5] ...

  3. javascript基础之javascript的存在形式和js代码块在页面中的存放位置

    1.存在形式 文件 如: <script src='js/jc.js'></script> 前页面 <script type='text/javascript'>a ...

  4. java 线程池用法

    public ThreadPoolExecutor(int corePoolSize, int maximumPoolSize, long keepAliveTime, TimeUnit unit, ...

  5. 如何打开Windows Server 2003 内存寻址扩展

    本文介绍了如何在系统内存大于4G的情况下,让windows2003 Advanced Server支持大内存的方法: 由于Windows2003 32bit是32位操作系统,当服务器配备内存高达4G时 ...

  6. 解压版mysql安装--windows系统

    1 解压到某个目录 2 配置配置文件 3 执行命令:安装目录/bin/mysqld --install mysql5.6 --defaults-file=指定配置文件位置 "安装目录/bin ...

  7. datagrid公用字段扩展

    easyui 1.3.5扩展 datagrid 控件 扩展公用段 把 (9082行) var opts = $.data(_698,"datagrid").options; 改成 ...

  8. IEnumerable中的 Any方法

    IEnumerable类中的 Any方法,表示集合中有任何一元素满足条件,返回就true , 该方法有两个重载 1. 不带任何参数,表示集合中有元素 2. 参入一个 Func<TSource, ...

  9. Installing Mp4box in centos 6

    Installing Mp4box in centos 6   Installing Mp4box in centos 6Login to the server cd /usr/local/src/ ...

  10. c++中构造函数 、析构函数的作用域详解

    我们知道,在c++中,析构函数是在函数作用域尾部执行析构函数,从而释放对象,但是有一种情况下,析构函数作用域发生变化,请看下面的例子,定义了一个Stock类,Stock类存放在stock.h中,主调用 ...