CF337 D. Book of Evil

题目链接：http://codeforces.com/problemset/problem/337/D
题意：有n个村庄，其中有m个村庄收到了魔鬼书本的影响，d是书本的最大影响距离，求这个书本可能在的村庄的个数。
题解：这道题就是在m个村庄里找到两个距离最长的村庄，然后分别以这两个村庄为基点求其与其他村庄的距离，两个距离都不大于d的即为可能存在的位置

#include<cstdio>
#include<cstring>
#include<vector>
using namespace std;
const int maxn = 1e5 + 5;
int d1[maxn];
int d2[maxn];
int p[maxn];
vector<int>graph[maxn];
void dfs(int node,int father,int deep, int *d)
{
    d[node] = deep;
    for(int i = 0; i < graph[node].size(); i ++){
        int tmp = graph[node][i];
        if(tmp!=father){
            dfs(tmp,node,deep+1,d);
        }
    }
}
int main()
{
    int n,m,d;
    while(~scanf("%d%d%d",&n,&m,&d)){
        memset(d1,0,sizeof(d1));
        memset(d2,0,sizeof(d2));
        memset(p,0,sizeof(p));
        memset(graph,0,sizeof(graph));
        for(int i = 1; i <= m; i ++){
            scanf("%d",&p[i]);
        }
        int a, b;
        for(int i = 1; i <= n - 1; i ++){
            scanf("%d%d",&a,&b);
            graph[a].push_back(b);
            graph[b].push_back(a);
        }
        dfs(1,0,0,d1);
        int u = 0,max_u = 0;
        for(int i = 1; i <= m; i ++){
            if(max_u<d1[p[i]]){
                max_u = d1[p[i]];
                u = p[i];
            }
        }
        dfs(u,0,0,d1);
        int v = 0,max_v = 0;
         for(int i = 1; i <= m; i ++){
            if(max_v<d1[p[i]]){
                max_v = d1[p[i]];
                v = p[i];
            }
        }
        dfs(v,0,0,d2);
        int cnt = 0;
        for(int i = 1; i <= n; i ++){
            if(d1[i] <= d && d2[i] <= d){
                cnt++;
            }
        }
        printf("%d\n",cnt);
    }
    return 0;
}