树形DP Codeforces Round #135 (Div. 2) D. Choosing Capital for Treeland

 

题目传送门

/*
    题意：求一个点为根节点，使得到其他所有点的距离最短，是有向边，反向的距离+1
    树形DP：首先假设1为根节点，自下而上计算dp[1](根节点到其他点的距离)，然后再从1开始，自上而下计算dp[v],
            此时可以从上个节点的信息递推出来
*/
#include <cstdio>
#include <cstring>
#include <cmath>
#include <vector>
using namespace std;

const int MAXN = 2e5 + 10;
const int INF = 0x3f3f3f3f;
struct Edge    {
    int v, w;
};
vector<Edge> G[MAXN];
int dp[MAXN];
bool vis[MAXN];
int n, res;

void DFS(int u)    {
    vis[u] = true;
    for (int i=0; i<G[u].size (); ++i)    {
        int v = G[u][i].v, w = G[u][i].w;
        if (vis[v])    continue;
        DFS (v);
        res += w;
    }
}

void DFS2(int u)    {
    vis[u] = true;
    for (int i=0; i<G[u].size (); ++i)    {
        int v = G[u][i].v, w = G[u][i].w;
        if (vis[v])    continue;
        if (w == 0)    dp[v] = dp[u] + 1;
        else    dp[v] = dp[u] - 1;
        DFS2 (v);
    }
}

void work(void)    {
    memset (vis, false, sizeof (vis));
    memset (dp, 0, sizeof (dp));

    res = 0;    DFS (1);    dp[1] = res;
    memset (vis, false, sizeof (vis));
    DFS2 (1);

    int mn = INF, p = 0;
    for (int i=1; i<=n; ++i)    {
        if (mn >= dp[i])    {
            mn = dp[i];    p = i;
        }
    }
    printf ("%d\n", mn);
    for (int i=1; i<=n; ++i)    {
        if (i == p)    {
            printf ("%d\n", i);    break;
        }
        if (dp[i] == mn)    printf ("%d ", i);
    }
}

int main(void)    {        //Codeforces Round #135 (Div. 2) D. Choosing Capital for Treeland
//    freopen ("A.in", "r", stdin);

    while (scanf ("%d", &n) == 1)    {
        for (int i=1; i<=n; ++i)    G[i].clear ();
        for (int i=1; i<=n-1; ++i)    {
            int u, v;    scanf ("%d%d", &u, &v);
            G[u].push_back ((Edge) {v, 0});
            G[v].push_back ((Edge) {u, 1});
        }

        work ();
    }

    return 0;
}