题目链接:http://codeforces.com/problemset/problem/219/D
题意:一张有向图上有n个点n-1条边,现在要求把这个图上的几个边的方向倒置,使得所有点都能到达同一个点。求最小需要翻转多少条边
把这张图转换成带权的无向图,原始的方向上权值是1,加一条反向边权值是0。这样就转换成了求一个点,使得从这个点出发到达所有点的权值和最大的问题了。
先预处理以点1为树根的各点到其子树各点的权值和,再算以其他点为根,到其他各点的权值和。即翻转当前节点和其父亲节点相连的那条边,并把父亲到各个节点的距离加上(要减去父亲到自己的边权)。
1 /* 2 ━━━━━┒ギリギリ♂ eye! 3 ┓┏┓┏┓┃キリキリ♂ mind! 4 ┛┗┛┗┛┃\○/ 5 ┓┏┓┏┓┃ / 6 ┛┗┛┗┛┃ノ) 7 ┓┏┓┏┓┃ 8 ┛┗┛┗┛┃ 9 ┓┏┓┏┓┃ 10 ┛┗┛┗┛┃ 11 ┓┏┓┏┓┃ 12 ┛┗┛┗┛┃ 13 ┓┏┓┏┓┃ 14 ┃┃┃┃┃┃ 15 ┻┻┻┻┻┻ 16 */ 17 #include <algorithm> 18 #include <iostream> 19 #include <iomanip> 20 #include <cstring> 21 #include <climits> 22 #include <complex> 23 #include <fstream> 24 #include <cassert> 25 #include <cstdio> 26 #include <bitset> 27 #include <vector> 28 #include <deque> 29 #include <queue> 30 #include <stack> 31 #include <ctime> 32 #include <set> 33 #include <map> 34 #include <cmath> 35 using namespace std; 36 #define fr first 37 #define sc second 38 #define cl clear 39 #define BUG puts("here!!!") 40 #define W(a) while(a--) 41 #define pb(a) push_back(a) 42 #define Rint(a) scanf("%d", &a) 43 #define Rs(a) scanf("%s", a) 44 #define Cin(a) cin >> a 45 #define FRead() freopen("in", "r", stdin) 46 #define FWrite() freopen("out", "w", stdout) 47 #define Rep(i, len) for(int i = 0; i < (len); i++) 48 #define For(i, a, len) for(int i = (a); i < (len); i++) 49 #define Cls(a) memset((a), 0, sizeof(a)) 50 #define Clr(a, x) memset((a), (x), sizeof(a)) 51 #define Full(a) memset((a), 0x7f7f, sizeof(a)) 52 #define lp p << 1 53 #define rp p << 1 | 1 54 #define pi 3.14159265359 55 #define RT return 56 #define lowbit(x) x & (-x) 57 #define onenum(x) __builtin_popcount(x) 58 typedef long long LL; 59 typedef long double LD; 60 typedef unsigned long long ULL; 61 typedef pair<int, int> pii; 62 typedef pair<string, int> psi; 63 typedef map<string, int> msi; 64 typedef vector<int> vi; 65 typedef vector<LL> vl; 66 typedef vector<vl> vvl; 67 typedef vector<bool> vb; 68 69 const int maxn = 200200; 70 int n, m; 71 vector<pii> G[maxn]; 72 int dp1[maxn], dp2[maxn]; 73 74 void dfs1(int u, int p) { 75 Rep(i, G[u].size()) { 76 int v = G[u][i].first; 77 int w = G[u][i].second; 78 if(v == p) continue; 79 dfs1(v, u); 80 dp1[u] += dp1[v] + w; 81 } 82 } 83 84 void dfs2(int u, int p) { 85 Rep(i, G[u].size()) { 86 int v = G[u][i].first; 87 int w = G[u][i].second; 88 if(v == p) continue; 89 dp2[v] += dp2[u] + dp1[u] - dp1[v] - w + !w; 90 dfs2(v, u); 91 } 92 } 93 94 bool cmp(pii a, pii b) { 95 if(a.first == b.first) return a.second < b.second; 96 return a.first > b.first; 97 } 98 99 int main() { 100 // FRead(); 101 int u, v; 102 while(~Rint(n)) { 103 For(i, 1, n+1) G[i].clear(); 104 Cls(dp1); Cls(dp2); 105 Rep(i, n-1) { 106 Rint(u); Rint(v); 107 G[u].push_back(pii(v, 1)); 108 G[v].push_back(pii(u, 0)); 109 } 110 dfs1(1, 0); dfs2(1, 0); 111 vector<pii> ret; 112 pii tmp; 113 For(i, 1, n+1) { 114 tmp.first = dp1[i] + dp2[i]; 115 tmp.second = i; 116 ret.push_back(tmp); 117 } 118 sort(ret.begin(), ret.end(), cmp); 119 int ans = n - 1 - ret[0].first; 120 int cnt = 1; 121 For(i, 1, ret.size()) { 122 if(ret[i].first == ret[0].first) cnt++; 123 } 124 printf("%d ", ans); 125 Rep(i, cnt) printf("%d ", ret[i].second); 126 printf(" "); 127 } 128 RT 0; 129 }