CodeForces 219D Choosing Capital for Treeland (树形DP)

题意：给一个n节点的有向无环图，要找一个这样的点：该点到其它n-1要逆转的道路最少。

析：很明显的树形DP，两次dfs，对于边，进行处理，如果是正向就是1，反向是-1，先进行dfs，计算出向子结点的方向要反转几条边，然后再第二次考虑，从父结点和子结点考虑。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e16;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-6;
const int maxn = 200000 + 10;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r > 0 && r <= n && c > 0 && c <= m;
}

struct Edge{
  int to, next, val;
};
Edge edge[maxn<<1];
int head[maxn], cnt;
int in[maxn];
int ans[maxn], res;
int dp[maxn];

void add_edge(int u, int v, int val){
  edge[cnt].to = v;
  edge[cnt].val = val;
  edge[cnt].next = head[u];
  head[u] = cnt++;
}

void dfs1(int u, int fa){
  dp[u] = 0;
  for(int i = head[u]; ~i; i = edge[i].next){
    int v = edge[i].to;
    if(v == fa)  continue;
    dfs1(v, u);
    if(edge[i].val == 1)  dp[u] += dp[v];
    else dp[u] += dp[v] + 1;
  }
}

void dfs2(int u, int fa, int fv){
  ans[u] = dp[u] + fv;
  res = min(res, ans[u]);
  for(int i = head[u]; ~i; i = edge[i].next){
    int v = edge[i].to;
    if(v == fa)  continue;
    dfs2(v, u, fv+dp[u]-dp[v]+edge[i].val);
  }
}

int main(){
  scanf("%d", &n);
  memset(head, -1, sizeof head);
  for(int i = 1; i < n; ++i){
    int u, v;
    scanf("%d %d", &u, &v);
    add_edge(u, v, 1);
    add_edge(v, u, -1);
  }
  dfs1(1, -1);
  res = n - 1;
  dfs2(1, -1, 0);
  printf("%d
", res);
  for(int i = 1; i <= n; ++i)
    if(res == ans[i])  printf("%d ", i);
  printf("
");
  return 0;
}