Poj 3352-edge dual-connection

The question requires at least several sides to be added to make the graph still connected after removing people and one edge

First, perform a dual-connection deflation point. A connected block is equivalent to a point in the new graph.

The new graph is a tree.

If you want to remove any edge in an original image after adding several edges, the image is still connected. We should connect the leaves so that any leaf has edges connected to other leaves.

Obviously, when the total number of leaves CNT is an even number, the answer is CNT/2;

When the total number of leaves is odd, the answer is (CNT + !) /2;

View code

# Include <stdio. h>
# Include <queue>
# Include < String . H>
# Include <vector>
 Using   Namespace STD;
 Const   Int Maxn =5005 ;
 Struct Edge
{
 Int S, T;
 Int Next;
 Int Vis;
} Edge [ 1000005 ];
 Int Head [maxn];
 Int E;
 Void Add_edge ( Int S,Int T)
{
Edge [e]. S = s;
Edge [e]. t = T;
Edge [e]. vis = 0 ;
Edge [e]. Next = head [s];
Head [s] = e ++;
}
 Int Btype, time;
 Int Dfn [maxn], low [maxn], belong [maxn];
 Int St [maxn], top;
Inline Int Min ( Int A,Int B ){ Return A <B? A: B ;}
 Void DFS ( Int S)
{
 Int I, T, ID;
St [++ top] = s;
Dfn [s] = low [s] = ++ time;
 For (I = head [s]; I! =- 1 ; I = edge [I]. Next)
{
 If (Edge [I]. Vis) Continue ;
Edge [I]. vis = edge [I ^ 1 ]. Vis = 1 ;
T = edge [I]. t;
 If (! Dfn [T])
{
DFS (t );
Low [s] = min (low [s], low [T]);
}
 Else Low [s] = min (low [s], dfn [T]);
}
 If (Dfn [s] = low [s])
{
Btype ++;
 Do {
T = sT [top --];
Belong [T] = btype;
} While (T! = S );
}
}
 Void SCC ( Int N)
{
 Int I;
Time = 0 ; Btype = 0 ; Top = 0 ;
Memset (dfn, 0 ,Sizeof ( Int ) * (N + 1 ));
 For (I = 1 ; I <= N; I ++) If (! Dfn [I])
DFS (I );
}
 Int Chu [maxn];
 Int N, m;
 Int Ma [ 1010 ] [ 1010 ];
Int Main ()
{
 Int I, J, K;
 Int A, B;
 While (Scanf ( "  % D  " , & N, & M )! = EOF)
{
Memset (Head ,- 1 , Sizeof (Head); E = 0 ;
Memset (CHU,0 , Sizeof (CHU ));
 For (I = 0 ; I <m; I ++)
{
Scanf ( "  % D  " , & A, & B );
Add_edge (A, B );
Add_edge (B, );
}
SCC (N );
Memset (MA, 0 , Sizeof (Ma ));
 For (I = 1 ; I <= N; I ++)
{
 For (J = head [I]; J! =- 1 ; J = edge [J]. Next)
{
 Int V = edge [J]. t;
 If (Belong [I]! = Belong [v] &! Ma [belong [I] [belong [v])
{
Ma [belong [I] [belong [v] = ma [belong [v] [belong [I] = 1 ;
Chu [belong [I] ++;
Chu [belong [v] ++;
}
}
}
 Int Ans = 0 ;
 For (I = 1 ; I <= btype; I ++)
{
 If (Chu [I] = 1 )
Ans ++;
}
If (ANS % 2 = 0 ) Ans/= 2 ;
 Else Ans = (ANS + 1 )/ 2 ;
Printf ( "  % D \ n  " , ANS );
}
 Return   0 ;
}