hdu 4587 TWO NODES (枚举+割点，5级)

TWO NODES


Time Limit: 24000/12000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 547    Accepted Submission(s): 179




Problem Description
Suppose that G is an undirected graph, and the value of stab is defined as follows:




Among the expression,G-i, -j is the remainder after removing node i, node j and all edges that are directly relevant to the previous two nodes. cntCompent is the number of connected components of X independently.
Thus, given a certain undirected graph G, you are supposed to calculating the value of stab.
 


Input
The input will contain the description of several graphs. For each graph, the description consist of an integer N for the number of nodes, an integer M for the number of edges, and M pairs of integers for edges (3<=N,M<=5000).
Please note that the endpoints of edge is marked in the range of [0,N-1], and input cases ends with EOF.
 


Output
For each graph in the input, you should output the value of stab.
 


Sample Input
4 5
0 1
1 2
2 3
3 0
0 2
 


Sample Output
2
 


Source
2013 ACM-ICPC南京赛区全国邀请赛——题目重现
 


Recommend

思路：枚举一个点，然后跑割点的连通块数。

#include<cstring>
#include<cstdio>
#include<iostream>
#define FOR(i,a,b) for(int i=a;i<=b;++i)
#define clr(f,z) memset(f,z,sizeof f)
using namespace std;
const int mp=5009;
const int me=mp+mp;


class Edge
{
public:
  int v,next;
};


class Graph_tree
{
public:
  int head[mp],edge;
  int block[mp],dfn[mp],dfs_clock;
  bool mark[mp];
  Edge e[me];
  int n;


  void clear()
  {
    clr(head,-1);edge=0;
  }


  void add(int u,int v)
  {
    e[edge].v=v;e[edge].next=head[u];head[u]=edge++;
  }


  int tarjan(int u,int rt)
  {
    int lowu,lowv,v;
    lowu=dfn[u]=++dfs_clock;
    mark[u]=1;
    for(int i=head[u];~i;i=e[i].next)
    {
      v=e[i].v;
      if(v==rt)continue;
      if(!dfn[v])
      {
        lowv=tarjan(v,rt);
        lowu=min(lowu,lowv);
        if(lowv>=dfn[u])
          ++block[u];
      }
      else if(mark[v])
        lowu=min(lowu,dfn[v]);
    }
    return lowu;
  }


  void getans()
  {
    int ans=0;
    FOR(i,0,n-1)///the point erase
    {
      clr(dfn,0);clr(mark,0);//clr(block,0);
      int num=0;dfs_clock=0;
      FOR(j,0,n-1)
      block[j]=1;///if no root block shuld add root
      FOR(j,0,n-1)
      if(i!=j&&!mark[j])
      {
        ++num;
        block[j]=0;
        tarjan(j,i);
      //  cout<<num<<" "<<block[j]<<endl;
      }
      FOR(j,0,n-1)
      if(i!=j)
        ans=max(ans,num+block[j]-1);
    }
    printf("%d\n",ans);
  }
}sp;


int main()
{
  int n,m,a,b;
  while(~scanf("%d%d",&n,&m))
  {
    sp.clear();
    sp.n=n;
    FOR(i,0,m-1)
    {
      scanf("%d%d",&a,&b);
      sp.add(a,b);sp.add(b,a);
    }
    //puts("++");
    sp.getans();
  }
  return 0;
}