CodeForces 219D Choosing Capit

题目链接：http://codeforces.com/contest/219/problem/D

题目大意：

　　给定一个n个节点的数和连接n个节点的n - 1条有向边，现在要选定一个节点作为起始节点，从这个点出发需要能走到其余每个节点，途中必然要调整有向边的方向，请求出当选定哪些节点作为初始节点时，所要调整的有向边最少，输出最小调整的边数和这些节点。

分析：

　　Hint：城市结构是树型结构。

代码如下：

#include <bits/stdc++.h>
using namespace std;
 
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define For(i,s,t) for (int i = (s); i <= (t); ++i)
#define rFor(i,t,s) for (int i = (t); i >= (s); --i)
#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
#define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
 
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
 
#define LOWBIT(x) ((x)&(-x))
 
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
 
#define ms0(a) memset(a,0,sizeof(a))
#define msI(a) memset(a,inf,sizeof(a))
#define msM(a) memset(a,-1,sizeof(a))
 
#define pii pair<int,int> 
#define piii pair<pair<int,int>,int> 
#define mp make_pair
#define pb push_back
#define fi first
#define se second
 
inline int gc(){
    static const int BUF = 1e7;
    static char buf[BUF], *bg = buf + BUF, *ed = bg;
     
    if(bg == ed) fread(bg = buf, 1, BUF, stdin);
    return *bg++;
} 
 
inline int ri(){
    int x = 0, f = 1, c = gc();
    for(; c<48||c>57; f = c=='-'?-1:f, c=gc());
    for(; c>47&&c<58; x = x*10 + c - 48, c=gc());
    return x*f;
}
 
typedef long long LL;
typedef unsigned long long uLL;
const int inf = 1e9 + 9;
const LL mod = 1e9 + 7;
const int maxN = 2e5 + 7;
 
struct Node{
    vector< pii > next;
};
 
int n, ans;
// dp[i]表示以i号城市为首都所要反转的道路数
int dp[maxN];
int vis[maxN];
Node nodes[maxN];
queue< int > Q;
 
// 求dp[x]
inline int dfs(int x) {
    if(nodes[x].next.empty()) return 0;
     
    int ret = 0;
    foreach(i, nodes[x].next) {
        int y = i->fi, z = i->se;
        if(vis[y]) continue;
        vis[x] = 1;
        if(z == -1) ++ret;
        ret += dfs(y);
    }
    return ret;
}
 
// 如果dp[i]已知，假设j号城市与i号城市相连，由于整个城市图是树形结构，
// 因此dp[j]与dp[i]除了他们的连线有分歧，其余应该反转的道路数量都是一样的
// 因此可由已算出来的1号城市为中心，向外做BFS
inline void bfs() {
    ms0(vis);
    vis[1] = 1;
    Q.push(1);
    ans = dp[1];
 
    while(!Q.empty()) {
        int tmp = Q.front();
        Q.pop();
 
        foreach(i, nodes[tmp].next) {
            int y = i->fi, z = i->se;
            if(vis[y]) continue;
            vis[y] = 1;
            dp[y] = dp[tmp] + z;
            ans = min(ans, dp[y]);
            Q.push(y);
        }
    }
}
 
int main(){
    while(cin >> n) {
        ms0(vis);
        rep(i, n + 1) nodes[i].next.clear();
        rep(i, n-1) {
            int x, y;
            cin >> x >> y;
            nodes[x].next.push_back(mp(y, 1));
            nodes[y].next.push_back(mp(x, -1));
        }
        dp[1] = dfs(1);
         
        bfs();
         
        cout << ans << endl;
        For(i, 1, n) if(ans == dp[i]) cout << i << " ";
        cout << endl;
    }
    return 0;
}