Codeforces 337 D Book of Evil(樹形dp，兩遍dfs)

題目鏈接：
Codeforces 337 D Book of Evil
題意：
給一個n 個節點和n−1 條邊的樹，有m 個給定的點，求這n 個點中到這m 個點的距離都小於等於d 的點的個數？樹上兩點的距離就是樹上兩點間路徑的邊的數量。
數據範圍：1≤m≤n≤105,0≤d≤n−1 。
分析：
需要一步轉化。我們求每個點到這m 個點的最遠距離，只要最遠距離大於d ，那麼這個點就是不可取的。這樣子一來就很簡單了，兩遍dfs ，記錄距離最大值和次大值即可。

#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <math.h>
using namespace std;
typedef long long ll;
const int MAX_N = 100010;

int n, m, d, total;
int head[MAX_N], id[MAX_N], flag[MAX_N];
int down[MAX_N], ddown[MAX_N];

struct Edge {
    int v, next;
} edge[MAX_N * 2];

void AddEdge(int u, int v)
{
    edge[total].v = v;
    edge[total].next = head[u];
    head[u] = total++;
}

void init()
{
    total = 0;
    for (int i = 0; i <= n; ++i) {
        head[i] = -1;
        down[i] = ddown[i] = id[i] = flag[i] = 0;
    }
}

void dfs_son(int u, int p)
{
    for (int i = head[u]; i != -1; i = edge[i].next) {
        int v = edge[i].v;
        if (v == p) continue;
        dfs_son(v, u);
        int son = down[v];
        if (flag[v] || down[v] > 0) son = down[v] + 1;
        if (son > down[u]) {
            ddown[u] = down[u];
            down[u] = son;
            id[u] = v;
        } else if (son > ddown[u]) {
            ddown[u] = son;
        }
    }
}

void dfs_father(int u, int p)
{
    for (int i = head[u]; i != -1; i = edge[i].next) {
        int v = edge[i].v;
        if (v == p) continue;

        int father;
        if (id[u] != v) father = down[u];
        else father = ddown[u];
        if (flag[u] || father > 0) father++;
        if (father > down[v]) {
            ddown[v] = down[v];
            down[v] = father;
            id[v] = u;
        } else if (father > ddown[v]) {
            ddown[v] = father;
        }
        dfs_father(v, u);
    }
}

int main()
{
    while (~scanf("%d%d%d", &n, &m, &d)) {
        init();
        for (int i = 0; i < m; ++i) {
            int u;
            scanf("%d", &u);
            flag[u] = 1;
        }
        for (int i = 1; i < n; ++i) {
            int u, v;
            scanf("%d%d", &u, &v);
            AddEdge(u, v);
            AddEdge(v, u);
        }
        dfs_son(1, 0);
        dfs_father(1, 0);
        int ans = n;
        for (int i = 1; i <= n; ++i) {
            if (down[i] > d) ans--;
        }
        printf("%d\n", ans);
    }
    return 0;
}