Network stream vertex connectivity poj 1966 zoj 2182

Vertex connectivity: Removes at least a few vertices to make the graph not connected.

Suppose two neighboring vertices u and v;

Two paths from u to V without common internal vertices are called independent tracks. The maximum number of independent paths from u to V is recorded as P (u, v)

Therefore, if the graph is not connected, you should remove a vertex on each independent track (not any). Therefore, the vertex connectivity is the minimum value of the maximum number of independent tracks.

Note: If the graph is a full graph, remove all vertices.

How can we find the maximum independent Rail P (A, B) between two points?

1. Split each point into two points, and connect an edge u '--- U "with a capacity of 1. E = (u, v) there are two arcs e '= <u ", V'> and E" = <v ", U'> in the new network. The two sides have an infinite capacity.

2. Set "A" as the Source Vertex and "B" as the sink vertex.

3. Find the maximum stream between two points

The sum of all the arc flows out of A is P (A, B), and all the arcs with flow 1 (the arc at this time is equivalent to a point) the corresponding vertex forms a cut-top set. After removing these vertices in the graph, A and B are disconnected.

I think twice, and after data verification, the book should be wrong. We need to enumerate the Source and Sink points in pairs.

Because if the enumerated source point is the point in the cut-top set, it will not work.

5 6)

The answer should be 1.

However, if there are no two-byte enumeration source points, the answer is 5.

Therefore, we still need to enumerate the two vertices.

View code

# Include <stdio. h>
# Include < String . H>
 Const   Int Max = 100005 ;
 Const   Int INF = 1000000000 ;

 Struct 
{
 Int V, C, next;
} Edge [ 1000000 ];
 Int E, head [Max];
 Int Gap [Max], cur [Max];
 Int Pre [Max], DIS [Max];
 Void Add_edge ( Int S, Int T, Int C, Int CC)
{
 /*  When adding an edge, add two at the same time,
One is positive, one is opposite,
Generally, the reverse capacity is 0.  */ 
Edge [e]. V = T; edge [e]. C = C;
Edge [e]. Next = head [s];
Head [s] = e ++;
Edge [e]. V = s; edge [e]. c = cc;
Edge [e]. Next = head [T];
Head [T] = e ++;
}
 Int Min ( Int A, Int B ){ Return (A =- 1 | B <)? B: ;}
 Int SAP ( Int S, Int T, Int N)
{}
Void Init ()
{
 For ( Int I = 0 ; I <E; I + = 2 )
{
Edge [I]. C + = edge [I ^ 1 ]. C;
Edge [I ^ 1 ]. C = 0 ;
}
}
 Int Main ()
{
Int I, J;
 Int A, B, n, m;
 While (Scanf ( "  % D  " , & N, & M )! = EOF)
{
Memset (Head ,- 1 , Sizeof (Head); E = 0 ;
 For (I = 0 ; I <n; I ++)
{
Add_edge (I, I + N, 1 , 0 );
}
 For (I = 0 ; I <m; I ++)
{
Scanf ( "  (% D, % d)  " , & A, & B );
Add_edge (a + N, B, INF, 0 );
Add_edge (B + N, A, INF, 0 );
}
 Int Ans = inf;
 For (I = 0 ; I <n; I ++)
{
 For (J = 0 ; J <n; j ++)
{
 If (J! = I ){
 Int TMP = SAP (I + N, J, 2 * N); Init ();
Ans = TMP <ans? TMP: ans;
}
}
}
 If (ANS> = inf) ans = N;
Printf ( "  % D \ n  " , ANS );
}
 Return   0 ;
}