poj1523 解题报告

题意：给出一个无向图，要求找出每个割点，以及除去割点之后能把图分成几个连通块

题解：点的双联通，统计割点属于多少个连通块

代码：

[cpp]
#include
#include
#include
using namespace std ;

const int MAXN = 1005 ;

struct type
{
int end , next ;
} ;

type edge[MAXN*MAXN] ;
int head[MAXN] , Count , top ;
int low[MAXN] , dfn[MAXN] , stack[MAXN] , vis[MAXN] , ans[MAXN] ; // ans是每个点属于的集合的数量，大于1就是割点了

void addedge( int start , int end )
{
edge[Count].end = end ;
edge[Count].next = head[start] ;
head[start] = Count ++ ;
}

void tarjan_dfs( int cur , int father , int &cnt , int &Time )
{
dfn[cur] = low[cur] = Time ++ ;
vis[cur] = 1 ;
for( int i = head[cur] ; i != -1 ; i = edge[i].next )
{
int end = edge[i].end ;
if( end == father ) continue ;
if( !vis[end] )
{
tarjan_dfs( end , cur , cnt , Time ) ;
low[cur] = min( low[cur] , low[end] ) ;
if( dfn[cur] <= low[cur] ) ans[cur] ++ ; // dfn = low 也可以
}
else low[cur] = min( low[cur] , dfn[end] ) ;
}
}

int tarjan( int N )
{
int Time = 0 , cnt = 0 ;
top = 0 ;
memset( dfn , 0 , sizeof( dfn ) ) ;
memset( low , 0 , sizeof( low ) ) ;
memset( vis , 0 , sizeof( vis ) ) ;
tarjan_dfs( 1 , -1 , cnt , Time ) ;
return cnt ;
}

int main()
{
int i , j , Max ;
int start , end , cnt ;

memset( head , -1 , sizeof( head ) ) ;
Count = 0 , cnt = 1 ;

while( scanf( "%d" , & start ) && start )
{
memset( head , -1 , sizeof( head ) ) ;
Count = 0 ;
Max = 0 ;
if( start > Max ) Max = start ;
scanf( "%d" , & end ) ;
if( end > Max ) Max = end ;
addedge( start , end ) ;
addedge( end , start ) ;
while( scanf( "%d" , & start ) && start )
{
if( start > Max ) Max = start ;
scanf( "%d" , & end ) ;
if( end > Max ) Max = end ;
addedge( start , end ) ;
addedge( end , start ) ;
}

for( i = 2 ; i <= Max ; i ++ ) ans[i] = 1 ;
ans[1] = 0 ;

tarjan( Max ) ;

printf( "Network #%d\n" , cnt ) ;
int flag = 0 ;
for( i = 1 ; i <= Max ; i ++ )
if( ans[i] > 1 )
{
printf( " SPF node %d leaves %d subnets\n" , i , ans[i] ) ;
flag = 1 ;
}
if( !flag ) printf( " No SPF nodes\n" ) ;
printf( "\n" ) ;
cnt ++ ;
}
return 0 ;