牡丹江.2014B(图论，树的直径）

 1 //#pragma comment(linker, "/STACK:1024000000,1024000000") 
 2 using namespace std ;
 3 const int M = 2e5 + 10 , inf = 111111111 ;
 4 struct node {
 5         int L , R ;
 6         int dl , dr ;
 7         node () {}
 8         node (int L , int R , int dl , int dr) :
 9                 L(L) , R(R) , dl(dl) , dr(dr) {}
10 }a[M] ;
11 int id , maxn ;
12 int n ;
13 vector<int> g[M] ;
14 bool vis[M] ;
15 vector<int> path ;
16 int N ;
17 int dfs (int o , int u) {
18         a[u] = node (u , u , 1 , 1) ;
19         for (vector<int>::iterator v = g[u].begin () ; v != g[u].end () ; v ++) {
20                 if (*v == o) continue ;
21                 int tmp = 1 + dfs (u , *v) ;
22                 if (tmp > a[u].dl) a[u] = node (a[*v].L , a[u].L , tmp , a[u].dl) ;
23                 else if (tmp > a[u].dr) a[u].R = a[*v].L , a[u].dr = tmp ;
24         }
25         if (a[u].dl + a[u].dr > maxn) maxn = a[u].dl + a[u].dr , id = u ;
26         return a[u].dl ;
27 }
28 
29 bool PATH (int o , int u) {
30         for (vector<int>::iterator v = g[u].begin() ; v != g[u].end () ; v ++) {
31                 if (*v == o) continue ;
32                 if (PATH (u , *v)) {path.push_back (u) ; return true ;} 
33         }
34         if (u == a[id].R) {path.push_back (u) ; return true ;} 
35         return false ;
36 }
37 
38 void DFS (int o , int u , int dep , int deep) {
39         if (dep > deep) return ;
40         vis[u] = 1 ;
41         for (vector<int>::iterator v = g[u].begin() ; v != g[u].end () ; v ++) {
42                 if (*v == o) continue ;
43                 DFS (u , *v , dep+1 , deep) ;
44         }
45 }
46 
47 bool judge (int x) {
48         memset (vis , 0 , sizeof(vis)) ;
49         DFS (-1 , path[x] , 0 ,  x ) ;
50         DFS (-1 , N-x-1 == x ? path[x+1] : path[N-x-1] , 0 , x ) ;
51         //printf ("x = %d\n" , x ) ;
52         //if (x == 1) for (int i = 1 ; i <= n ; i ++) if (!vis[i]) printf ("geng shi %d\n" , i ) ;
53         for (int i = 1 ; i <= n ; i ++) if (!vis[i]) return false ;
54         return true ;
55 }
56 
57 void solve () {
58         N = path.size () ;
59         int l = 0 , r = N / 2 ;
60         int ret = r ;
61         //printf ("l = %d , r = %d\n" , l , r) ;
62         while (l <= r) {
63                 int mid = l + r >> 1 ;
64                 if (judge (mid)) {
65                         ret = mid ;
66                         r = mid - 1 ;
67                         //printf ("ret = %d\n" , ret ) ;
68                 }
69                 else l = mid + 1 ;
70         }
71         printf ("%d %d %d\n" , ret , path[ret] , ret == N-ret-1 ? path[ret+1] : path[N-ret-1] ) ;  
72 }
73 
74 int main () {
75         int T ;
76         scanf ("%d" , &T ) ;
77         while (T --) {
78                 scanf ("%d" , &n) ;
79                 path.clear () ;
80                 for (int i = 0 ; i <= n ; i ++) g[i].clear () ;
81                 for (int i = 0 ; i < n - 1 ; i ++) {
82                         int u , v ;
83                         scanf ("%d%d" , &u , &v) ;
84                         g[u].push_back (v) ;
85                         g[v].push_back (u) ;
86                 }
87 
88                 maxn = -1 ;
89                 dfs (-1 , 1) ;
90                 //printf ("Dia = %d , Lnode = %d , Rnode = %d\n" , maxn-1 , a[id].L , a[id].R) ;
91                 PATH (-1 , a[id].L) ;
92                 //for (vector<int>::iterator it = path.begin() ; it != path.end() ; it ++) printf ("%d " , *it) ; puts ("") ;
93                 solve () ;
94         }
95         return 0 ;
96 }