树形结构，挺有意思的题目。不难。
1 /* 219D */ 2 #include <iostream> 3 #include <string> 4 #include <map> 5 #include <queue> 6 #include <set> 7 #include <stack> 8 #include <vector> 9 #include <deque> 10 #include <algorithm> 11 #include <cstdio> 12 #include <cmath> 13 #include <ctime> 14 #include <cstring> 15 #include <climits> 16 #include <cctype> 17 #include <cassert> 18 #include <functional> 19 #include <iterator> 20 #include <iomanip> 21 using namespace std; 22 //#pragma comment(linker,"/STACK:102400000,1024000") 23 24 #define sti set<int> 25 #define stpii set<pair<int, int> > 26 #define mpii map<int,int> 27 #define vi vector<int> 28 #define pii pair<int,int> 29 #define vpii vector<pair<int,int> > 30 #define rep(i, a, n) for (int i=a;i<n;++i) 31 #define per(i, a, n) for (int i=n-1;i>=a;--i) 32 #define clr clear 33 #define pb push_back 34 #define mp make_pair 35 #define fir first 36 #define sec second 37 #define all(x) (x).begin(),(x).end() 38 #define SZ(x) ((int)(x).size()) 39 #define lson l, mid, rt<<1 40 #define rson mid+1, r, rt<<1|1 41 42 const int maxn = 2e5+5; 43 int D[maxn]; 44 int W[maxn], sw; 45 int W_[maxn]; 46 vpii E[maxn]; 47 vi vc; 48 int ans; 49 int n; 50 51 void dfs(int u, int fa, int w_, int d) { 52 int v, w; 53 54 D[u] = d; 55 W_[u] = w_; 56 rep(i, 0, SZ(E[u])) { 57 v = E[u][i].fir; 58 w = E[u][i].sec; 59 if (v == fa) 60 continue; 61 dfs(v, u, w_+w, d+1); 62 W[u] += W[v] + w; 63 } 64 } 65 66 void solve() { 67 // calculate w & c 68 dfs(1, -1, 0, 0); 69 ans = sw = W[1]; 70 vc.pb(1); 71 72 // calculate the minimum cost 73 int tmp; 74 75 rep(i, 2, n+1) { 76 tmp = sw - W_[i] + D[i] - W_[i]; 77 if (tmp == ans) { 78 vc.pb(i); 79 } else if (tmp < ans) { 80 vc.clear(); 81 ans = tmp; 82 vc.pb(i); 83 } 84 } 85 86 printf("%d ", ans); 87 rep(i, 0, SZ(vc)) 88 printf("%d ", vc[i]); 89 putchar(' '); 90 } 91 92 int main() { 93 ios::sync_with_stdio(false); 94 #ifndef ONLINE_JUDGE 95 freopen("data.in", "r", stdin); 96 freopen("data.out", "w", stdout); 97 #endif 98 99 int u, v; 100 101 scanf("%d", &n); 102 rep(i, 1, n) { 103 scanf("%d %d", &u, &v); 104 E[u].pb(mp(v, 0)); 105 E[v].pb(mp(u, 1)); 106 } 107 108 solve(); 109 110 #ifndef ONLINE_JUDGE 111 printf("time = %d. ", (int)clock()); 112 #endif 113 114 return 0; 115 }