SPF
| Time Limit: 1000MS | Memory Limit: 10000K | |
| Total Submissions: 5855 | Accepted: 2693 |
Description
the left would prevent some of the still available nodes from communicating with each other. Nodes 1 and 2 could still communicate with each other as could nodes 4 and 5, but communication between any other pairs of nodes would no longer be possible.
Node 3 is therefore a Single Point of Failure (SPF) for this network. Strictly, an SPF will be defined as any node that, if unavailable, would prevent at least one pair of available nodes from being able to communicate on what was previously a fully connected
network. Note that the network on the right has no such node; there is no SPF in the network. At least two machines must fail before there are any pairs of available nodes which cannot communicate.
Input
1 2 and 2 1 specify the same connection. All node numbers will range from 1 to 1000. A line containing a single zero ends the list of connected nodes. An empty network description flags the end of the input. Blank lines in the input file should be ignored.
Output
The first network in the file should be identified as "Network #1", the second as "Network #2", etc. For each SPF node, output a line, formatted as shown in the examples below, that identifies the node and the number of fully connected subnets that remain when
that node fails. If the network has no SPF nodes, simply output the text "No SPF nodes" instead of a list of SPF nodes.
Sample Input
1 2
5 4
3 1
3 2
3 4
3 5
0
1 2
2 3
3 4
4 5
5 1
0
1 2
2 3
3 4
4 6
6 3
2 5
5 1
0
0
Sample Output
Network #1
SPF node 3 leaves 2 subnets
Network #2
No SPF nodes
Network #3
SPF node 2 leaves 2 subnets
SPF node 3 leaves 2 subnets
Source
传送门：【POJ】1523 SPF
题目分析：求割点的模板题，tarjan一次得到所有的割点，然后对每个割点的儿子dfs一遍，dfs的次数就是删除该割点能得到的连通块的数量。
代码如下：
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std ;
#define REP( i , n ) for ( int i = 0 ; i < n ; ++ i )
#define REPF( i , a , b ) for ( int i = a ; i <= b ; ++ i )
#define REPV( i , a , b ) for ( int i = a ; i >= b ; -- i )
#define clear( a , x ) memset ( a , x , sizeof a )
const int MAXN = 1005 ;
const int MAXE = 1000000 ;
struct Edge {
int v , n ;
Edge ( int var = 0 , int next = 0 ) : v ( var ) , n ( next ) {}
} ;
struct BCC {
Edge edge[MAXE] ;
int adj[MAXN] , cntE ;
int cut[MAXN] ;
int low[MAXN] , dfn[MAXN] , dfs_clock ;
bool vis[MAXN] ;
void init () {
cntE = dfs_clock = 0 ;
clear ( cut , 0 ) ;
clear ( dfn , 0 ) ;
clear ( adj , -1 ) ;
}
void addedge ( int u , int v ) {
edge[cntE] = Edge ( v , adj[u] ) ;
adj[u] = cntE ++ ;
edge[cntE] = Edge ( u , adj[v] ) ;
adj[v] = cntE ++ ;
}
void tarjan ( int u , int fa ) {
low[u] = dfn[u] = ++ dfs_clock ;
int son = 0 ;
for ( int i = adj[u] ; ~i ; i = edge[i].n ) {
int v = edge[i].v ;
if ( !dfn[v] ) {
++ son ;
tarjan ( v , u ) ;
low[u] = min ( low[u] , low[v] ) ;
if ( low[v] >= dfn[u] )
cut[u] = 1 ;
}
else if ( v != fa )
low[u] = min ( low[u] , dfn[v] ) ;
}
if ( fa < 0 && son == 1 )
cut[u] = 0 ;
}
void dfs ( int u ) {
vis[u] = 1 ;
for ( int i = adj[u] ; ~i ; i = edge[i].n )
if ( !vis[edge[i].v] )
dfs ( edge[i].v ) ;
}
void solve () {
int flag = 0 ;
REP ( u , MAXN ) {
if ( cut[u] ) {
flag = 1 ;
int num = 0 ;
clear ( vis , 0 ) ;
vis[u] = 1 ;
for ( int i = adj[u] ; ~i ; i = edge[i].n ) {
int v = edge[i].v ;
if ( !vis[v] ) {
++ num ;
dfs ( v ) ;
}
}
printf ( " SPF node %d leaves %d subnets\n" , u , num ) ;
}
}
if ( !flag )
printf ( " No SPF nodes\n" ) ;
}
} ;
BCC c ;
int input () {
c.init () ;
int u , v ;
int flag = 0 ;
while ( scanf ( "%d" , &u ) ) {
if ( !u ) {
if ( !flag )
return 0 ;
else
return 1 ;
}
flag = 1 ;
scanf ( "%d" , &v ) ;
c.addedge ( u , v ) ;
}
return 0 ;
}
void work () {
int cas = 0 ;
while ( input () ) {
printf ( "Network #%d\n" , ++ cas ) ;
REP ( i , MAXN )
if ( ~c.adj[i] ) {
c.tarjan ( i , -1 ) ;
break ;
}
c.solve () ;
printf ( "\n" ) ;
}
}
int main () {
work () ;
return 0 ;
}