题意：存在有n个地方， 有两个人分别从1和n号地点出发，问在最短的时间里他们在哪个地点能能够相遇，如果有多个地点则输出多个节点。然后给了m个集合，每个集合中的所有地点之间距离为t。

题解：该题很容易想到最短路然后枚举所有可能相遇的点取max后取min，难点在于建图，如果暴力建图的话会出现n^2复杂度，所以我们把每个集合当成一个源点，源点到集合中点的距离为t，集合中点到该源点的距离为0，这样建图就是O(n)了。然后分别跑1和n的最短路，枚举各点即可。

AC代码：

#include <bits/stdc++.h>
using namespace std;
#define met(a, b) memset(a, b, sizeof(a))
#define rep(i, a, b) for(int i = a; i < b; i++)
#define per(i, a, b) for(int i = a; i >= b; i--)
#define fi first
#define se second
#define pb push_back
#define mp make_pair
typedef long long ll;
const int maxn = 1e6 + 10;
const int inf = 0x3f3f3f3f;
ll gcd(ll a, ll b) {
   return b == 0 ? a : gcd(b, a % b);}
ll lcm(ll a, ll b) {
   return a / gcd(a, b) * b;}
ll qPow(ll base, ll n) {ll res = 1; while(n) {
   if(n & 1) res *= base; base *= base; n >>= 1;} return res;}
int T, n, m, Head[maxn << 1], edgesNum = 0, t, s, d;
int dis[10][101000];
struct EDGE {
    int to, from, w, nex;
} edges[maxn << 2];
void init() {edgesNum = 0; met(Head, -1); for(int i = 0; i <= n + m; i++) dis[0][i] = dis[1][i] = inf;}
void addEdge(int u, int v, int w) {
    edges[edgesNum].from = u;
    edges[edgesNum].to = v;
    edges[edgesNum].w = w;
    edges[edgesNum].nex = Head[u];
    Head[u] = edgesNum++;
}
void spfa(int s);
int main() {
    int cas = 1;
    scanf("%d", &T);
    while(T--) {
        scanf("%d%d", &n, &m);
        init();
        rep(i, 1, m + 1) {
            scanf("%d%d", &t, &s);
            rep(j, 0, s) {
                scanf("%d", &d);
                addEdge(n + i, d, t);
                addEdge(d, n + i, 0);
            }
        }
        spfa(1);
        spfa(n);
        int MIN = inf;
        vector<int> v;
        v.clear();
        rep(i, 1, n + 1) {
//            printf("dis[%d] = %d\n", i, dis[0][i]);
//            printf("dis[%d] = %d\n", i, dis[1][i]);
            int t = max(dis[0][i], dis[1][i]);
            if(t < MIN && t != inf) {
                v.clear();
                MIN = t;
                v.pb(i);
            } else if(t == MIN && t != inf) {
                v.pb(i);
            }
        }
        //sort(v.begin(), v.end());
        int len = v.size();
        if(MIN != inf) {
            printf("Case #%d: %d\n", cas++, MIN);
            rep(i, 0, len) {
                printf("%d", v[i]);
                if(i != len - 1) printf(" ");
            }
            puts("");
        } else {
   printf("Case #%d: Evil John\n", cas++);}
    }
    return 0;
}
void spfa(int s) {
    int pos = (s == 1 ? 0 : 1);
    bool vis[maxn];
    met(vis, false);
    dis[pos][s] = 0;
    vis[s] = true;
    queue<int> q;
    while(!q.empty()) q.pop();
    q.push(s);
    while(!q.empty()) {
        int u = q.front();
        q.pop();
        vis[u] = false;
        for(int e = Head[u]; ~e; e = edges[e].nex) {
            int v = edges[e].to;
            if(dis[pos][v] > dis[pos][u] + edges[e].w) {
                dis[pos][v] = dis[pos][u] + edges[e].w;
                if(!vis[v]) {
                    q.push(v);
                    vis[v] = true;
                }
            }
        }
    }

}