pku3352 Road Construction

#include  < iostream >
using   namespace  std;

#define  MAXN 1001
#define  min(a,b) (a<b?a:b)

int  p[MAXN],n,ecnt,m,dfn[MAXN],lowlink[MAXN],sign,deg[MAXN],color[MAXN],totcol;

struct  Edge{
     int  v,next;
}edg[ 10 * MAXN];

void  init(){
    ecnt = 0 ;
    sign = 0 ;
    memset(p, - 1 , sizeof (p));
    memset(dfn, - 1 , sizeof (dfn));
    memset(deg, 0 , sizeof (deg));
    
    memset(color, 0 , sizeof (color));
    totcol = 0 ;
}

void  dfs( int  pre, int  u){
     int  i,v;
    dfn[u] = lowlink[u] =++ sign;
     for (i = p[u];i !=- 1 ;i = edg[i].next){
        v = edg[i].v;
         if (dfn[v] !=- 1 ){
             if (v != pre)
                lowlink[u] = min(lowlink[u],dfn[v]);
        }
         else {
            dfs(u,v);
            lowlink[u] = min(lowlink[u],lowlink[v]);
        }
    }
}

void  flood( int  u){
     int  i,v;
     for (i = p[u];i !=- 1 ;i = edg[i].next){
        v = edg[i].v;
         if ( ! color[v]){
             if (dfn[u] < lowlink[v])
                color[v] =++ totcol;
             else
                color[v] = color[u];
            flood(v);
        }
    }
}




int  solve(){
     int  i,u,v;
    dfs( - 1 , 1 );
    
    color[ 1 ] =++ totcol;
    flood( 1 ); // 着色区分图内各分量
    
     for (u = 1 ;u <= n;u ++ ){
         for (i = p[u];i !=- 1 ;i = edg[i].next){
            v = edg[i].v;
             if (color[u] != color[v])
                deg[color[u]] ++ ; // 由于图是由两条相反的有向边表示无向边的，计算某结点的度只需计算其出度
        }
    }
    
     int  ans = 0 ;
     for (i = 1 ;i <= totcol;i ++ )
         if (deg[i] == 1 ) // 叶子的度为1
            ans ++ ;
        
         return  ans;
}


int  main(){
     int  i,u,v;
     while (scanf( " %d%d " , & n, & m) != EOF){
        init();
         for (i = 0 ;i < m;i ++ ){
            scanf( " %d%d " , & u, & v);
            edg[ecnt].next = p[u];
            edg[ecnt].v = v;
            p[u] = ecnt ++ ;
            edg[ecnt].next = p[v];
            edg[ecnt].v = u;
            p[v] = ecnt ++ ;
        }
        printf( " %d\n " ,(solve() + 1 ) / 2 ); // 每添一条边，就选择树中的两个叶子结点，将它们连起来，于是最少的添边数目就是(叶子结点个数+1)/2
        
        
    }
     return   0 ;
}