poj 1523 求割点

题意：是求一个无向图的割点，和连通子图个数。

 

根据割点定义：

1. u为根， 则u至少有两个儿子。

2. u不为根，则至少存在某一儿子节点s, low[s] >= d[u], 即s和s的后代不会追溯到比u更早的祖先点。

 

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<iostream>
#include<vector>
#include<string>
#include<math.h>
#include<map>
#include<set>
#include<algorithm>
using namespace std;
#define MAXN 1000100
struct Edge
{
  int u, next;
  Edge( ) { }
  Edge( int U, int Next) : u(U), next(Next) { }
}edge[MAXN];

int head[MAXN], e, ptime, N;
int d[1010], low[1010], cut[1010], cnt;

void init( )
{
  memset(head, -1, sizeof(head));
  e = 0;
  ptime = 0;
  memset(d, 0, sizeof(d));
  memset(low,0,sizeof(low));
  memset(cut,0,sizeof(cut));
}

void AddEdge( int u, int v)
{
   edge[e]= Edge(v, head[u]);
   head[u] = e++;
   edge[e] = Edge(u, head[v]);
   head[v] = e++;
}

//找割点 
void find_cut( int u )
{
   low[u] = d[u] = ++ptime;
   int son = 0;
   for( int e = head[u]; e != -1; e = edge[e].next )
   {
      int v = edge[e].u;
      if( !d[v] )
      {
          son++;
          find_cut( v );
          if( (u != 1 && low[v] >= d[u]) || (u == 1) && son >= 2 )
              cut[u]++;
          low[u] = min(low[u], low[v]);
      }
      else
      {
         low[u] = min( low[u], d[v]);            
      }     
        
   }     
}

void solve( )
{ 
  find_cut(1);
  bool f = false;
  for( int i = 1; i <= N; i++)
  { 
    if( cut[i] )
    printf("  SPF node %d leaves %d subnets\n", i, cut[i] + 1), f = true;
  }     
  if( !f )
    puts("  No SPF nodes");
  puts("");
}

int main( )
{ 
   int a, b, abc = 1;
   while( scanf("%d", &a) != EOF )
   {
     init( ); 
     if( a == 0 ) break;
     scanf("%d",&b);
     while( 1 )
     {
       AddEdge(a, b);
       N = max(N, max(a,b));
       scanf("%d",&a);
       if( a == 0 )
         break;
       scanf("%d",&b);  
     }
     printf("Network #%d\n", abc++);
     solve( );
   }
   return 0;
}