UVA 318 Network(无向图求割点数)

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.

求割点数:

模板题:

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<string>
#include<iostream>
#include<queue>
#include<cmath>
#include<map>
#include<stack>
#include<bitset>
using namespace std;
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define CLEAR( a , x ) memset ( a , x , sizeof a )
typedef long long LL;
typedef pair<int,int>pil;
const int maxn=110;
const int maxm=110*110;
struct node{
    int to,next;
    bool col;//为桥
}e[maxm];
int head[maxn],cnt;
int DFN[maxn],low[maxn];
int s[maxn],instack[maxn];
int idex,top,bridge;
int col[maxn],add[maxn];//为割;删除点后增加的联通快
int n;
void init()
{
    cnt=top=idex=bridge=0;
    CLEAR(head,-1);
    CLEAR(DFN,0);
    CLEAR(low,0);
    CLEAR(instack,0);
    CLEAR(col,0);
    CLEAR(add,0);
}
void addedge(int u,int v)
{
    e[cnt].to=v;e[cnt].next=head[u];
    e[cnt].col=false;head[u]=cnt++;
}
void Tarjan(int u,int pre)
{
    low[u]=DFN[u]=++idex;

    instack[u]=1;
    int son=0;
    for(int i=head[u];i!=-1;i=e[i].next)
    {
        int v=e[i].to;
        if(!DFN[v])
        {
            son++;
            Tarjan(v,u);
            if(low[u]>low[v]) low[u]=low[v];
            if(low[v]>DFN[u])//桥
            {
                bridge++;
                e[i].col=true;
                e[i^1].col=true;
            }
            if(u!=pre&&low[v]>=DFN[u])//割点判断1
            {
                col[u]=1;
                add[u]++;
            }
        }
        else if(low[u]>DFN[v])
            low[u]=DFN[v];
    }
    if(u==pre&&son>1)  col[u]=1;//割点判断2
    if(u==pre)  add[u]=son-1;
    instack[u]=0;
    top--;
}
int main()
{
    int u,v;
    while(~scanf("%d",&n)&&n)
    {
        init();
        while(~scanf("%d",&u)&&u)
        {
            while(getchar()!='\n')
            {
                scanf("%d",&v);
                addedge(u,v);
                addedge(v,u);
            }
        }
        REPF(i,1,n)
           if(!DFN[i])  Tarjan(i,i);
        int ans=0;
        REPF(i,1,n)
          if(col[i])  ans++;
        printf("%d\n",ans);
    }
    return 0;
}


郑重声明:本站内容如果来自互联网及其他传播媒体,其版权均属原媒体及文章作者所有。转载目的在于传递更多信息及用于网络分享,并不代表本站赞同其观点和对其真实性负责,也不构成任何其他建议。