poj 1144 Network 无向图求割点

Network

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.

思路：就是求割点没什么说的；就是输入小心点

　　

#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define mod 1000000007
#define inf 999999999
int scan()
{
    int res = 0 , ch ;
    while( !( ( ch = getchar() ) >= '0' && ch <= '9' ) )
    {
        if( ch == EOF )  return 1 << 30 ;
    }
    res = ch - '0' ;
    while( ( ch = getchar() ) >= '0' && ch <= '9' )
        res = res * 10 + ( ch - '0' ) ;
    return res ;
}
int dfn[1010];
int low[1010];
int head[1010];
int ans[1010];
int visit[1010];
int index,root,son,node,jiedge;
struct is
{
    int u,v;
    int next;
}edge[1010];
void add(int x,int y)
{
    jiedge++;
    edge[jiedge].u=x;
    edge[jiedge].v=y;
    edge[jiedge].next=head[x];
    head[x]=jiedge;
    jiedge++;
    edge[jiedge].u=y;
    edge[jiedge].v=x;
    edge[jiedge].next=head[y];
    head[y]=jiedge;
}
void dfs(int u)
{
    for(int i=head[u];i;i=edge[i].next)
    {
        int v=edge[i].v;
        if(visit[v]==0)
        {
            visit[v]=1;++index;
            dfn[v]=low[v]=index;
            dfs(v);
            low[u]=min(low[u],low[v]);
            if(low[v]>=dfn[u])
            {
                if(u==root)
                son++;
                else
                ans[u]++;
            }
        }
        else
        low[u]=min(low[u],dfn[v]);
    }
}
void trajan()
{
    memset(visit,0,sizeof(visit));
    memset(ans,0,sizeof(ans));
    index=1;
    root=1;
    son=0;
    low[1]=dfn[1]=1;
    visit[1]=1;
    dfs(1);
}
int main()
{
    int u,v;
    int flag=0;
    while(scanf("%d",&node)!=EOF)
    {
        memset(head,0,sizeof(head));
        jiedge=0;
        if(node==0)break;
        while(1)
        {
            int u,v;
            scanf("%d",&u);
            if(!u)break;
            while(getchar()!='\n')
            {
                scanf("%d",&v);
                add(u,v);
            }
        }
        trajan();
        int answer=0;
        if(son>1)
        answer++;
        for(int i=2;i<=node;i++)
        if(ans[i])
        answer++;
        printf("%d\n",answer);
    }
    return 0;
}

View Code

 

转载于:https://www.cnblogs.com/jhz033/p/5380806.html