Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 8254   Accepted: 3772

Description

Consider the two networks shown below. Assuming that data moves around these networks only between directly connected nodes on a peer-to-peer basis, a failure of a single node, 3, in the network on the left would prevent some of the still available nodes from communicating with each other. Nodes 1 and 2 could still communicate with each other as could nodes 4 and 5, but communication between any other pairs of nodes would no longer be possible.

Node 3 is therefore a Single Point of Failure (SPF) for this network. Strictly, an SPF will be defined as any node that, if unavailable, would prevent at least one pair of available nodes from being able to communicate on what was previously a fully connected network. Note that the network on the right has no such node; there is no SPF in the network. At least two machines must fail before there are any pairs of available nodes which cannot communicate. 

Input

The input will contain the description of several networks. A network description will consist of pairs of integers, one pair per line, that identify connected nodes. Ordering of the pairs is irrelevant; 1 2 and 2 1 specify the same connection. All node numbers will range from 1 to 1000. A line containing a single zero ends the list of connected nodes. An empty network description flags the end of the input. Blank lines in the input file should be ignored.

Output

For each network in the input, you will output its number in the file, followed by a list of any SPF nodes that exist.

The first network in the file should be identified as "Network #1", the second as "Network #2", etc. For each SPF node, output a line, formatted as shown in the examples below, that identifies the node and the number of fully connected subnets that remain when that node fails. If the network has no SPF nodes, simply output the text "No SPF nodes" instead of a list of SPF nodes.

Sample Input

1 2
5 4
3 1
3 2
3 4
3 5
0 1 2
2 3
3 4
4 5
5 1
0 1 2
2 3
3 4
4 6
6 3
2 5
5 1
0 0

Sample Output

Network #1
SPF node 3 leaves 2 subnets Network #2
No SPF nodes Network #3
SPF node 2 leaves 2 subnets
SPF node 3 leaves 2 subnets

Source

点双连通分量模板题

 /*by SilverN*/
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<vector>
using namespace std;
const int mxn=;
struct edge{
int v,nxt;
}e[mxn<<];
int hd[mxn],mct=;
void add_edge(int u,int v){
e[++mct].v=v;e[mct].nxt=hd[u];hd[u]=mct;
return;
}
int n;
int low[mxn],dfn[mxn],dtime=;
int num[mxn];
int st[mxn],top=;
bool cut[mxn];
void DFS(int u,int fa){
low[u]=dfn[u]=++dtime;
st[++top]=u;
int i,j;
for(i=hd[u];i;i=e[i].nxt){
int v=e[i].v;
if(v==fa)continue;
if(!dfn[v]){
DFS(v,u);
if(low[v]>=dfn[u]){
cut[u]=;
int s=;
do{
s=st[top--];
++num[s];
}while(u!=s);
top++;
}
// num[u]++;
low[u]=min(low[u],low[v]);
}
else low[u]=min(dfn[v],low[u]);
}
return;
}
bool have[mxn];
int main(){
int cas=;
while(){
memset(hd,,sizeof hd);
memset(cut,,sizeof cut);
memset(low,,sizeof low);
memset(dfn,,sizeof dfn);
memset(have,,sizeof have);
memset(num,,sizeof num);
mct=dtime=top=;
int i,j,u,v;
bool flag=;
while(scanf("%d",&u) && u){
scanf("%d",&v);
have[u]=have[v]=;
add_edge(u,v);
add_edge(v,u);
flag=;
}
if(!flag)break;
for(i=;i<=;i++)
if(have[i] && !dfn[i])DFS(i,);
printf("Network #%d\n",++cas);
flag=;
for(i=;i<=;i++){
if(num[i]>=){
flag=;
printf(" SPF node %d leaves %d subnets\n",i,num[i]);
}
}
if(!flag)printf(" No SPF nodes\n");
printf("\n");
}
return ;
}

POJ1523 SPF的更多相关文章

  1. POJ1523 SPF[无向图割点]

    SPF Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 8139   Accepted: 3723 Description C ...

  2. POJ1523 SPF 单点故障

    POJ1523 题意很简单,求删除割点后原先割点所在的无向连通图被分成了几个连通部分(原题说prevent at least one pair of available nodes from bein ...

  3. POJ1523 SPF(割点模板)

    题目求一个无向图的所有割点,并输出删除这些割点后形成几个连通分量.用Tarjan算法: 一遍DFS,构造出一颗深度优先生成树,在原无向图中边分成了两种:树边(生成树上的边)和反祖边(非生成树上的边). ...

  4. POJ1523:SPF(无向连通图求割点)

    题目:http://poj.org/problem?id=1523 题目解析: 注意题目输入输入,防止PE,题目就是求割点,并问割点将这个连通图分成了几个子图,算是模版题吧. #include < ...

  5. poj图论解题报告索引

    最短路径: poj1125 - Stockbroker Grapevine(多源最短路径,floyd) poj1502 - MPI Maelstrom(单源最短路径,dijkstra,bellman- ...

  6. Tarjan求割点和桥

    by szTom 前置知识 邻接表存储及遍历图 tarjan求强连通分量 割点 割点的定义 在一个无向图中,如果有一个顶点集合,删除这个顶点集合以及这个集合中所有顶点相关联的边以后,图的连通分量增多, ...

  7. POJ1523:SPF——题解

    http://poj.org/problem?id=1523 这题明显就是求割点然后求割完之后的强连通分量的个数. 割点都会求,怎么求割完的分量个数呢? 我们可以通过万能的并查集啊!(具体做法看代码吧 ...

  8. POJ 1523 SPF(求割点)

    题目链接 题意 : 找出图中所有的割点,然后输出删掉他们之后还剩多少个连通分量. 思路 : v与u邻接,要么v是u的孩子,要么u是v的祖先,(u,v)构成一条回边. //poj1523 #includ ...

  9. POJ 1523 SPF 割点与桥的推断算法-Tarjan

    题目链接: POJ1523 题意: 问一个连通的网络中有多少个关节点,这些关节点分别能把网络分成几部分 题解: Tarjan 算法模板题 顺序遍历整个图,能够得到一棵生成树: 树边:可理解为在DFS过 ...

随机推荐

  1. iOS9 判断微信qq是否安装

    iOS 9检测QQ.微信是否安装无效的解决方法 在info.plist里面添加LSApplicationQueriesSchemes(Array类型),然后插入weixin, wechat, mqq的 ...

  2. javascript替换手机号中间4位

    // 匹配手机号首尾,以类似“123****8901”的形式输出 '12345678901'.replace(/(\d{3})\d{4}(\d{4})/, '$1****$2'); 此段正则匹配字符串 ...

  3. iOS之block

    1. Block的声明和线程安全Block属性的声明,首先需要用copy修饰符,因为只有copy后的Block才会在堆中,栈中的Block的生命周期是和栈绑定的,可以参考之前的文章(iOS: 非ARC ...

  4. Spring 4 bak

    IOC (参考<Spring企业开发>.<Spring实战 第三版  第四版>) IoC概述 1.           控制反转 2.依赖注入   控制反转:大多数情况下,想要 ...

  5. FormsAuthentication详解

    配置安全鉴别 鉴别是指鉴定来访用户是否合法的过程.ASP.NET Framework支持三种鉴别类型: Windows鉴别: NET Passport鉴别: Forms鉴别. 对于某一特定的应用程序, ...

  6. android 合并两个jar包

    你所要导出的类里边用到了别的jar包.比如说你写的类连接了数据库,用到数据库驱动包oracl.jar(也就是你导入到Myeclipse或eclipse的jdbc包).. .在dos环境下,进入到D盘的 ...

  7. cisco交换技术list

  8. C#中的默认访问修饰符

    1.命名空间下的元素的默认访问修饰符 public : 同一程序集的其他任何代码或引用该程序集的其他程序集都可以访问该类型或成员.internal : 同一程序集中的任何代码都可以访问该类型或成员,但 ...

  9. 供应商和管理员查看供应商地址簿信息SQL

    --管理员查看地址簿 SELECT hps.party_site_id, hps.party_site_name AS address_name, 'CURRENT' AS status, hzl.a ...

  10. ZooKeeper学习第八期——ZooKeeper伸缩性

    一.ZooKeeper中Observer 1.1 ZooKeeper角色 经过前面的介绍,我想大家都已经知道了在ZooKeeper集群当中有两种角色Leader和Follower.Leader可以接受 ...