| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 4247 | Accepted: 1988 |
Description
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
by one space. Each block ends with a line containing just 0. The last block has only one line with N = 0;
Output
Sample Input
5 5 1 2 3 4 0 6 2 1 3 5 4 6 2 0 0
Sample Output
1 2
Hint
Source
#include<cstdio>
#include<cstring>
const int V=110;
const int E=110*110;
int n;
struct EDGE
{
int v,next;
}edge[E];
int head[V],dep[V],low[V],cut[V],e;
bool vis[V];
void addedge(int u,int v)
{
edge[e].v=v;
edge[e].next=head[u];
head[u]=e++;
}
inline int min(int a,int b)
{
if(a>b) return b;
return a;
}
void dfs(int idx,int fa,int d)
{
dep[idx]=low[idx]=d;
vis[idx]=1;
for(int p=head[idx];p!=-1;p=edge[p].next)
{
int v=edge[p].v;
if(v==fa) continue;
if(vis[v])
{
low[idx]=min(low[idx],dep[v]);
continue;
}
dfs(v,idx,d+1);
low[idx]=min(low[idx],low[v]);
if(low[v]>=d) cut[idx]++;
}
if(fa!=-1) cut[idx]++;
}
int main()
{
int a,b;
char c;
while(scanf("%d",&n)&&n)
{
e=0;
memset(head,-1,sizeof(head));
memset(dep,0,sizeof(dep));
memset(low,0,sizeof(low));
memset(cut,0,sizeof(cut));
memset(vis,false,sizeof(vis));
while(scanf("%d",&a)!=EOF&&a)
{
while(scanf("%c",&c)!=EOF)
{
if(c=='/n') break;
scanf("%d",&b);
addedge(a,b);
addedge(b,a);
}
}
dfs(1,-1,0);
int sum=0;
for(int i=1;i<=n;i++)
if(cut[i]>1) sum++;
printf("%d/n",sum);
}
return 0;
}
本文介绍了一种算法,用于识别电话网络中导致其他节点失去连接的关键节点。通过构建网络模型并运用深度优先搜索等技术,文章详细解释了如何确定这些关键点。
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