CF 196 Div 2 D (树形dp)

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在一个由n个定居点构成的沼泽区域中，帕拉丁·马诺卡追踪到了一本古老的邪恶之书。通过已知受其影响的m个定居点及损害范围d，利用图论与算法知识确定可能藏匿该书的位置。

Paladin Manao caught the trail of the ancient Book of Evil in a swampy area. This area contains n settlements numbered from 1 to n. Moving through the swamp is very difficult, so people tramped exactly n - 1 paths. Each of these paths connects some pair of settlements and is bidirectional. Moreover, it is possible to reach any settlement from any other one by traversing one or several paths.

The distance between two settlements is the minimum number of paths that have to be crossed to get from one settlement to the other one. Manao knows that the Book of Evil has got a damage range d. This means that if the Book of Evil is located in some settlement, its damage (for example, emergence of ghosts and werewolves) affects other settlements at distance d or less from the settlement where the Book resides.

Manao has heard of m settlements affected by the Book of Evil. Their numbers are p1, p2, ..., pm. Note that the Book may be affecting other settlements as well, but this has not been detected yet. Manao wants to determine which settlements may contain the Book. Help him with this difficult task.

Input

The first line contains three space-separated integers n, m and d (1 ≤ m ≤ n ≤ 100000; 0 ≤ d ≤ n - 1). The second line contains mdistinct space-separated integers p1, p2, ..., pm (1 ≤ pi ≤ n). Then n - 1 lines follow, each line describes a path made in the area. A path is described by a pair of space-separated integers ai and bi representing the ends of this path.

Output

Print a single number — the number of settlements that may contain the Book of Evil. It is possible that Manao received some controversial information and there is no settlement that may contain the Book. In such case, print 0.

Sample test(s)

input
6 2 3
1 2
1 5
2 3
3 4
4 5
5 6

output
3

Note

Sample 1. The damage range of the Book of Evil equals 3 and its effects have been noticed in settlements 1 and 2. Thus, it can be in settlements 3, 4 or 5.

$\text{[math]}$

/*
思路:求出离i这个点最远的受影响的点作为i这点的属性，然后当这个属性小于等于d，那么这个点可能有鬼

*/

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
#include<stack>
#include<vector>
#include<set>
#include<map>

#define L(x) (x<<1)
#define R(x) (x<<1|1)
#define MID(x,y) ((x+y)>>1)

#define bug printf("hihi\n")

#define eps 1e-8
typedef __int64 ll;

using namespace std;
#define INF 0x3f3f3f3f
#define N 100005

int vis[N];
int n,m,d;
int head[N];
int num;

int dp[N][2];
int dir[N][2];

struct stud{
  int to,ne;
}e[N*2];

inline void add(int u,int v)
{
    e[num].to=v;
    e[num].ne=head[u];
    head[u]=num++;
}

void dfs(int u,int pre)
{
     if(vis[u]) dp[u][0]=dp[u][1]=0;
     else dp[u][0]=dp[u][1]=-INF;
     dir[u][0]=dir[u][1]=0;

    for(int i=head[u];i!=-1;i=e[i].ne)
    {
        int to=e[i].to;
        if(to==pre) continue;
        dfs(to,u);
        if(dp[to][1]+1>dp[u][0])
        {
            dp[u][0]=dp[to][1]+1;
            dir[u][0]=to;
            if(dp[u][0]>dp[u][1])
            {
                swap(dp[u][0],dp[u][1]);
                swap(dir[u][0],dir[u][1]);
            }
        }
    }
}

void dfss(int u,int pre)
{
     int i,j;
     for(int i=head[u];i!=-1;i=e[i].ne)
     {
         int to=e[i].to;
         if(to==pre) continue;
         if(dir[u][1]==to)
         {
             if(dp[u][0]+1>dp[to][0])
             {
                 dp[to][0]=dp[u][0]+1;
                 dir[to][0]=u;
                 if(dp[to][0]>dp[to][1])
                 {
                     swap(dp[to][0],dp[to][1]);
                     swap(dir[to][1],dir[to][0]);
                 }
             }
         }
         else
             if(dp[u][1]+1>dp[to][0])
             {
                 dp[to][0]=dp[u][1]+1;
                 dir[to][0]=u;
                 if(dp[to][0]>dp[to][1])
                 {
                     swap(dp[to][0],dp[to][1]);
                     swap(dir[to][1],dir[to][0]);
                 }
             }
        dfss(to,u);
     }
}

int main()
{
    int i,j;
    while(~scanf("%d%d%d",&n,&m,&d))
    {
        memset(head,-1,sizeof(head));
        num=0;
        int x;
        memset(vis,0,sizeof(vis));
        for(i=0;i<m;i++)
        {
            scanf("%d",&x);
            vis[x]=1;
        }
        int u,v;
        i=n-1;
        while(i--)
        {
            scanf("%d%d",&u,&v);
            add(u,v);
            add(v,u);
        }
       dfs(1,-1);

       int ans=0;
//       for(i=1;i<=n;i++)
//         printf("%d ",dp[i][1]);
//       printf("\n");
        dfss(1,-1);
       for(i=1;i<=n;i++)
        if(dp[i][1]<=d) ans++;

       printf("%d\n",ans);
    }
    return 0;
}