Network
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 10907   Accepted: 5042

Description

A Telephone Line Company (TLC) is establishing a new telephone cable network. They are connecting several places numbered by integers from 1 to N . No two places have the same number. The lines are bidirectional and always connect together two places and in each place the lines end in a telephone exchange. There is one telephone exchange in each place. From each place it is
possible to reach through lines every other place, however it need
not be a direct connection, it can go through several exchanges. From
time to time the power supply fails at a place and then the exchange
does not operate. The officials from TLC realized that in such a case it
can happen that besides the fact that the place with the failure is
unreachable, this can also cause that some other places cannot connect
to each other. In such a case we will say the place (where the failure

occured) is critical. Now the officials are trying to write a
program for finding the number of all such critical places. Help them.

Input

The
input file consists of several blocks of lines. Each block describes
one network. In the first line of each block there is the number of
places N < 100. Each of the next at most N lines contains the number
of a place followed by the numbers of some places to which there is a
direct line from this place. These at most N lines completely describe
the network, i.e., each direct connection of two places in the network
is contained at least in one row. All numbers in one line are separated

by one space. Each block ends with a line containing just 0. The last block has only one line with N = 0;

Output

The output contains for each block except the last in the input file one line containing the number of critical places.

Sample Input

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

Sample Output

1
2

Hint

You need to determine the end of one line.In order to make it's easy to determine,there are no extra blank before the end of each line.

Source

模板~

 #include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#define ll long long
using namespace std;
const int MAXN = ;
const int MAXM = ;
struct Edge
{
int to,next;
bool cut;
} edge[MAXM];
int head[MAXN],tot;
int Low[MAXN],DFN[MAXN],Stack[MAXN],Belong[MAXN];
int Index,top;
bool Instack[MAXN],cut[MAXN];
int bridge,add_block[MAXN]; void addedge(int u,int v)
{
edge[tot].to = v;
edge[tot].next = head[u];
edge[tot].cut = false;
head[u] = tot++;
}
void Tarjan(int u,int pre)
{
int v;
Low[u] = DFN[u] = ++Index;
Stack[top++] = u;
Instack[u] = true;
int son = ;
for(int i = head[u]; i != -; i = edge[i].next)
{
v = edge[i].to;
if(v == pre)
continue;
if( !DFN[v] )
{
son++;
Tarjan(v,u);
if(Low[u] > Low[v]) Low[u] = Low[v]; if(Low[v] > DFN[u])
{
bridge++;
edge[i].cut = true;
edge[i^].cut = true;
} if(u != pre && Low[v] >= DFN[u])
{
cut[u] = true;
add_block[u]++;
}
}
else if(Instack[v] && Low[u] > DFN[v])
Low[u] = DFN[v];
}
if(u == pre && son > ) cut[u] = true;
if(u == pre) add_block[u] = son -;
Instack[u] = false;
top--;
}
void solve(int N)
{
memset(DFN,,sizeof(DFN));
memset(Instack,false,sizeof(Instack));
memset(add_block,,sizeof(add_block));
memset(cut,false,sizeof(cut));
Index = top = bridge = ;
for(int i = ; i <= N; i++)
if( !DFN[i])
Tarjan(i,i);
int ans = ;
for(int i = ; i <= N; i++)
if(cut[i])
ans++;
printf("%d\n",ans);
}
void init()
{
tot = ;
memset(head,-,sizeof(head));
}
int main(void)
{
int n;
int maze[][];
while(scanf("%d",&n),n)
{
char ss[];
init();
memset(maze,,sizeof(maze));
getchar();
while(cin.getline(ss,),ss[] != '')
{
int u = ,cur = ,l = (int)strlen(ss);
while(ss[cur] != ' ' && cur < l)
{
u*= ;
u += ss[cur] - '';
cur++;
}
int v;
while(ss[cur] && cur < l)
{
v = ;
while(ss[cur] != ' ' && cur < l)
{
v*= ;
v += ss[cur] - '';
cur++;
}
maze[u][v] = maze[v][u] = ;
cur++;
}
}
for(int i = ; i <= n; i++)
for(int j = ; j < i; j++)
if(maze[i][j])
{
addedge(i,j);
addedge(j,i);
}
solve(n);
}
return ;
}

poj 1144 Network(割点 入门)的更多相关文章

  1. poj 1144 Network(割点)

    题目链接: http://poj.org/problem?id=1144 思路分析:该问题要求求出无向联通图中的割点数目,使用Tarjan算法即可求出无向联通图中的所有的割点,算法复杂度为O(|V| ...

  2. POJ 1144 Network(无向图连通分量求割点)

    题目地址:id=1144">POJ 1144 求割点.推断一个点是否是割点有两种推断情况: 假设u为割点,当且仅当满足以下的1条 1.假设u为树根,那么u必须有多于1棵子树 2.假设u ...

  3. POJ 1144 Network(Tarjan求割点)

    Network Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 12707   Accepted: 5835 Descript ...

  4. poj 1144 Network 图的割顶判断模板

    Network Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 8797   Accepted: 4116 Descripti ...

  5. poj 1144 Network(无向图求割顶数)

    题目链接:poj 1144 题意就是说有 n(标号为 1 ~ n)个网点连接成的一个网络,critical places 表示删去后使得图不连通的顶点,也就是割顶,求图中割顶的个数. 直接上大白书上的 ...

  6. poj 1144 Network 【求一个网络的割点的个数 矩阵建图+模板应用】

    题目地址:http://poj.org/problem?id=1144 题目:输入一个n,代表有n个节点(如果n==0就结束程序运行). 在当下n的这一组数据,可能会有若干行数据,每行先输入一个节点a ...

  7. poj 1144 Network【双连通分量求割点总数】

    Network Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 11042   Accepted: 5100 Descript ...

  8. POJ 1144 Network(tarjan 求割点个数)

    Network Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 17016   Accepted: 7635 Descript ...

  9. poj 1144 Network 无向图求割点

    Network Description A Telephone Line Company (TLC) is establishing a new telephone cable network. Th ...

随机推荐

  1. Luogu P4180 【模板】严格次小生成树[BJWC2010]

    P4180 [模板]严格次小生成树[BJWC2010] 题意 题目描述 小\(C\)最近学了很多最小生成树的算法,\(Prim\)算法.\(Kurskal\)算法.消圈算法等等.正当小\(C\)洋洋得 ...

  2. 3.pycharm spark配置

        pycharm 内的环境变量配置     选择相应的spark程序文件的对应的配置信息       PYSPARK_PYTHON:python的安装路径   PYTHONPATH:spark安 ...

  3. 转载:Linux命令经典面试题:统计文件中出现次数最多的前10个单词

    1.使用linux命令或者shell实现:文件words存放英文单词,格式为每行一个英文单词(单词可以重复),统计这个文件中出现次数最多的前10个单词 主要考察对sort.uniq命令的使用,相关解释 ...

  4. 廖雪峰Java11多线程编程-3高级concurrent包-5Atomic

    Atomic java.util.concurrent.atomic提供了一组原子类型操作: 如AtomicInteger提供了 int addAndGet(int delta) int increm ...

  5. DISTINCT 方法用于返回唯一不同的值 。

    DISTINCT 方法用于返回唯一不同的值 . 例如: $Model->distinct(true)->field('name')->select(); 生成的SQL语句是: SEL ...

  6. 莫烦PyTorch学习笔记(六)——批处理

    1.要点 Torch 中提供了一种帮你整理你的数据结构的好东西, 叫做 DataLoader, 我们能用它来包装自己的数据, 进行批训练. 而且批训练可以有很多种途径. 2.DataLoader Da ...

  7. Xcode导航栏功能简介

    1.Xcode 1.1.AboutXcode 1.2.Preferences General  Accounts   Behaviors1 Behavior2    Navigation Fonts& ...

  8. 从0开始学习ssh之日志工具与配置log4j

    添加slf4j-api-1.6.1,slf4j-log4j12-1.6.1,log4j-1.2.15三个jar包到lib文件夹下就可以使用log4j日志文件.具体配置在log4j.properties ...

  9. elasticsearch filters特性

    使用filters优化查询 ElasticSearch支持多种不同类型的查询方式,这一点大家应该都已熟知.但是在选择哪个文档应该匹配成功,哪个文档应该呈现给用户这一需求上,查询并不是唯一的选择.Ela ...

  10. tp5.1 swoole 实现异步处理

    客户端请求:<?phpnamespace app\index\controller; class Index{ public function index() { $client = new \ ...