二次聯通門 : luogu P2018 消息傳遞

/*
    luogu P2018 消息傳遞

    樹形dp
    原來用優先隊列做了一下， T了倆點
    MMP

    去看正解。。
    復雜度一樣好不好。。
    每次到達一個點，記錄其子樹中所有的dp值
    優先向大的一邊轉移
*/
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>

#define INF 1e8
const int BUF = 12312323
 
;
char Buf[BUF], *buf = Buf;

inline void read (int &now)
{
    for (now = 0; !isdigit (*buf); ++ buf);
    for (; isdigit (*buf); now = now * 10 + *buf - ‘0‘, ++ buf);
}

#define Max 1100

struct E
{
    E *n;
    int to;
};
E *list[Max], poor[Max << 1], *Tail = poor;
int father[Max], dp[Max];

inline 
 
int max (int a, int b)
{
    return a > b ? a : b;
}
inline int min (int a, int b)
{
    return a < b ? a : b;
}
void Dp (int now, int Father)
{
    int data[Max], C = 0;
    father[now] = Father;
    for (E *e = list[now]; e; e = e->n)
        if (e->to != Father)
        {
            Dp (e
 
->to, now);
            data[++ C] = dp[e->to];    
        }
    std :: sort (data + 1, data + 1 + C);
    for (int i = 1; i <= C; ++ i)
        dp[now] = max (dp[now], data[i] + C - i + 1);

}
int Answer[Max];

int Main ()
{
    fread (buf, 1, BUF, stdin); int Maxn = INF;
    int N, M; read (N); register int i; int x;
    for (i = 2; i <= N; ++ i)
    {
        read (x);
        ++ Tail, Tail->to = i, Tail->n = list[x], list[x] = Tail;
        ++ Tail, Tail->to = x, Tail->n = list[i], list[i] = Tail;
    }
    for (i = 1; i <= N; ++ i)
    {
        Dp (i, 0);
        Answer[i] = dp[i];
        Maxn = min (Maxn, Answer[i]);
        memset (dp, 0, sizeof dp);
    }
    printf ("%d\n", Maxn + 1);
    for (i = 1; i <= N; ++ i)
        if (Answer[i] == Maxn) printf ("%d ", i);    
    return 0;
}
int ZlycerQan = Main ();
int main (int argc, char *argv[]) {;}

luogu P2018 消息傳遞