| Network |
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 mostN 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 用来测试模板的 这里输入用了strtok的技巧
#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <string.h>
using namespace std; /*
* 求 无向图的割点和桥
* 可以找出割点和桥，求删掉每个点后增加的连通块。
* 需要注意重边的处理，可以先用矩阵存，再转邻接表，或者进行判重
*/
const int MAXN = ;
const int MAXM = ;
struct Edge
{
int to,next;
bool cut;//是否为桥的标记
}edge[MAXM];
int head[MAXN],tot;
int Low[MAXN],DFN[MAXN],Stack[MAXN];
int Index,top;
bool Instack[MAXN];
bool cut[MAXN];
int add_block[MAXN];//删除一个点后增加的连通块
int bridge; 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];
//桥
//一条无向边(u,v)是桥，当且仅当(u,v)为树枝边，且满足DFS(u)<Low(v)。
if(Low[v] > DFN[u])
{
bridge++;
edge[i].cut = true;
edge[i^].cut = true;
}
//割点
//一个顶点u是割点，当且仅当满足(1)或(2) (1) u为树根，且u有多于一个子树。
//(2) u不为树根，且满足存在(u,v)为树枝边(或称父子边，
//即u为v在搜索树中的父亲)，使得DFS(u)<=Low(v)
if(u != pre && Low[v] >= DFN[u])//不是树根
{
cut[u] = true;
add_block[u]++;
}
}
else if( Low[u] > DFN[v])
Low[u] = DFN[v];
}
//树根，分支数大于1
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 g[][];
char buf[];
int main()
{
int n;
while(scanf("%d",&n)== && n)
{
gets(buf);
memset(g,,sizeof(g));
while(gets(buf))
{
if(strcmp(buf,"")==)break;
char *p = strtok(buf," ");
int u;
sscanf(p,"%d",&u);
p = strtok(NULL," ");
int v;
while(p)
{
sscanf(p,"%d",&v);
p = strtok(NULL," ");
g[u][v]=g[v][u]=;
}
}
init();
for(int i = ;i <= n;i++)
for(int j = i+;j <= n;j++)
if(g[i][j])
{
addedge(i,j);
addedge(j,i);
}
solve(n);
}
return ;
}