POJ 1144 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

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

Central Europe 1996

The question is a template question, there is nothing to say, but I have not passed the fourth level, and it took a lot of effort to understand the input of this question.

The input for this question is .

Multiple sets of inputs, for each set

The first line is n, which means there are n points.

There are at most n lines next, representing the paths that exist, and for each line

The first number indicates the start point, the rest of the line indicates the end point

If in a group, there is only one 0 in a row, it means the end of the group of data.

If the input n is 0, it means that the input is all over.

Here is the AC code:

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 using namespace std;
 5 int root;
 6 int num[1000];
 7 int low[1000],book[1024];
 8 int Num;
 9 int n,ui,flag=0;
10 int u[1000],v[1000];
11 int first[1024],Next[10086];
12 int index,ans;
13 void dfs(int cur,int father)
14 {
15     int child=0;
16     index++;
17     num[cur]=index;
18     low[cur]=index;
19     int k=first[cur];
20     while(k!=-1){
21         if(num[v[k]]==0){
22             child++;
23             dfs(v[k],cur);
24             low[cur]=min(low[cur],low[v[k]]);
25             if(cur!=root&&low[v[k]]>=num[cur]){
26                 if(!book[cur]){ans++;}
27                 book[cur]=1;
28             }
29             if(cur==root&&child>=2){
30                 if(!book[cur]){ans++;}
31                 book[cur]=1;
32             }
33         }
34         else if(v[k]!=father){
35             low[cur]=min(low[cur],num[v[k]]);
36         }
37         k=Next[k];
38     }
39 }
40 
41 int main()
42 {
43     char c;
44     while(cin>>n&&n){
45         memset(num,0,sizeof(num));
46         memset(low,0,sizeof(low));
47         memset(book,0,sizeof(book));
48         ans=index=0;
49         for(int i=1;i<=n;i++){
50             first[i]=-1;
51         }
52         getchar();
53         flag=Num=0;
54         int Count=1;
55         while(~scanf("%c",&c)){
56             if(c!=10&&c!=' '){Num=Num*10+c-48;}
57             else {
58 
59                 if(Num==0&&flag==0){break;}
60                 if(!flag){ui=Num;}
61                 else{u[Count]=ui;v[Count]=Num;Next[Count]=first[ui];first[ui]=Count++;}
62                 if(c==10){flag=0;Num=0;continue;}
63                 Num=0;flag=1;
64             }
65         }
66         Count--;
67         for(int i=1;i<=Count;i++){
68             u[i+Count]=v[i];
69             v[i+Count]=u[i];
70             Next[i+Count]=first[v[i]];
71             first[v[i]]=i+Count;
72         }
73 
74 
75         root=1;
76         dfs(1,root);
77         cout<<ans<<endl;
78 
79 
80     }
81 }

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