337D Book of Evil

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一棵n个点的树，上面有m个ghost，每个ghost有距离为d的影响区域。求一共有多少个点，能被所有ghost影响到。

只要求出树上距离最远的两个ghost，如果一个点能同时被这两个ghost影响到，那一定能被suoyoughost影响到了！求树上距离最远的两个ghost，用类似于求树的直径的方法dfs就行了。

#include<algorithm>#include<iostream>#include<cstring>#include<cstdlib>#include<fstream>#include<sstream>#include<bitset>#include<vector>#include<string>#include<cstdio>#include<cmath>#include<stack>#include<queue>#include<stack>#include<map>#include<set>#define FF(i, a, b) for(int i=a; i<b; i++)#define FD(i, a, b) for(int i=a; i>=b; i--)#define REP(i, n) for(int i=0; i<n; i++)#define CLR(a, b) memset(a, b, sizeof(a))#define debug puts("**debug**")#define LL long long#define PB push_backusing namespace std;const int maxn = 100001;int n, m, x, y, z;int flag[maxn], dist[3][maxn];vector<int> G[maxn];void dfs(int u, int fa, int id, int d){    dist[id][u] = d;    REP(i, G[u].size())    {        int v = G[u][i];        if(v != fa) dfs(v, u, id, d+1);    }}int main(){    while(~scanf("%d%d%d", &n, &m, &z))    {        CLR(flag, 0);  REP(i, n) G[i].clear();        while(m--)        {            scanf("%d", &x);            flag[x-1] = 1;        }        REP(i, n-1)        {            scanf("%d%d", &x, &y);            G[--x].PB(--y); G[y].PB(x);        }        REP(i, n) if(flag[i])        {            x = i;            dfs(i, -1, 0, 0);            break;        }        REP(i, n) if(flag[i] && dist[0][i] > dist[0][x]) x = i;    dfs(x, -1, 1, 0);        REP(i, n) if(flag[i] && dist[1][i] > dist[1][x]) x = i;    dfs(x, -1, 2, 0);        int ans = 0;        REP(i, n) if(dist[1][i] <= z && dist[2][i] <= z) ans++;        printf("%d\n", ans);    }    return 0;}