poj1144 tarjan asks for cutting point naked question

Network

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 11684		Accepted: 5422

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

---

//tarjan算法求割点模板
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <string>
#include <algorithm>
#include <stack>
#include <queue>
#include <vector>

const int inf = 0x3f3f3f;
const int MAXN = 1e2+10;

using namespace std;

vector < int > adj[MAXN];
 int low[MAXN]; // Can access the first ancestor node position 
int dfn[MAXN]; // Record the node position 
int cur[MAXN]; // Record whether it is a cut point 
int rt,rt_sons; // The root node, the number of subtrees of the root node 
int n;

void init(){
    rt_sons = 0;
    rt = 1;
    memset(low,0,sizeof(low));
    memset(dfn,0,sizeof(dfn));
    memset(cur,0,sizeof(cur));
    for(int i=1;i<=n;i++)
        adj[i].clear();
}

void dfs(int u,int degree){
    dfn[u] = low[u] = degree;
    int v;
    for(int i=0;i<adj[u].size();i++){
        v = adj[u][i];
         if (!dfn[v]){ // The successor node has not been visited, indicating that a ring is not formed 
            dfs(v,degree+ 1 );
             if (u==rt)rt_sons++ ;
             else {
                 if (low[v]>=dfn[u]) // The successor node cannot find a node earlier than this node, it means that the node is a cut point 
                    cur[u] = 1 ;
                low[u] = min(low[u],low[v]); // Pay attention to updating the position of the node that can search for the parent node at all times, the latter node may form a ring, and the search is earlier than the node location 
            }
        }
        else { // The successor node has been visited, that is, in the case of loopback, the position that can be accessed must be updated 
            low[u] = min(low[u],dfn[v]);
        }

    }
}

intmain ()
{
    int u,v;
    while(scanf("%d",&n),n){
        init();
        while(scanf("%d",&u),u){
            while(getchar()!='\n'){
                scanf("%d",&v);
                adj[u].push_back(v);
                adj[v].push_back(u);
            }
        }
        dfs(rt, 1 );
         int sum= 0 ;
         if (rt_sons>= 2 )cur[ 1 ] = 1 ; // y has two subtrees whose roots are cut points 
        for ( int i= 1 ;i<=n; i++ ){
             if (cur[i])sum++ ;
        }
        /*cout<<"debug"<<endl;
        for(int i=1;i<=n;i++){
            cout<<cur[i]<<" ";
        }
        cout<<endl;
        cout<<"debug"<<endl;*/
        cout<<sum<<endl;
    }
    //cout << "Hello world!" << endl;
    return 0;
}

View Code

 

Reprinted in: https://www.cnblogs.com/EdsonLin/p/5447763.html