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Sol

走到一個點的前提是所有的影響它的點都走到，並且它的時間為那些點的時間\(max\)與自己的最短路的\(max\)
考慮\(Dijkstra\)
每次松弛時候，把它影響到的點的防護罩減\(1\)
如果某個點沒有後繼影響它的節點就丟到大根堆內
然後這個繼續松弛

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
const int _(3005);
const int
 
 __(70005);
typedef long long ll;

IL int Input(){
    RG int x = 0, z = 1; RG char c = getchar();
    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3
 
) + (c ^ 48);
    return x * z;
}

int n, first[_], cnt, m, d[_], vis[_];
ll dis[_], mxd[_];
struct Data{
    int u; ll d;

    IL int operator <(RG Data B) const{
        return d > B.d;
    }
};
struct Edge{
    int to, next, w;
} edge[__];
vector <int> G[_];
priority_queue <Data> Q;

IL void
 
 Add(RG int u, RG int v, RG int w){
    edge[cnt] = (Edge){v, first[u], w}, first[u] = cnt++;
}

IL ll Solve(){
    Fill(dis, 127), mxd[1] = dis[1] = 0, Q.push((Data){1, 0});
    while(!Q.empty()){
        RG Data x = Q.top(); Q.pop();
        if(vis[x.u]) continue;
        vis[x.u] = 1;
        for(RG int l = G[x.u].size(), i = 0; i < l; ++i){
            RG int v = G[x.u][i];
            mxd[v] = max(mxd[v], x.d);
            if(!--d[v]) Q.push((Data){v, max(mxd[v], dis[v])});
        }
        for(RG int e = first[x.u]; e != -1; e = edge[e].next){
            RG int v = edge[e].to, w = edge[e].w;
            if(vis[v]) continue;
            if(x.d + w < dis[v]){
                dis[v] = x.d + w;
                if(!d[v]) Q.push((Data){v, max(mxd[v], dis[v])});
            }
        }
    }
    return max(dis[n], mxd[n]);
}

int main(RG int argc, RG char* argv[]){
    Fill(first, -1), n = Input(), m = Input();
    for(RG int i = 1; i <= m; ++i){
        RG int u = Input(), v = Input(), w = Input();
        Add(u, v, w);
    }
    for(RG int i = 1; i <= n; ++i){
        d[i] = Input();
        for(RG int j = 1; j <= d[i]; ++j) G[Input()].push_back(i);
    }
    printf("%lld\n", Solve());
    return 0;
}

Bzoj1922: [Sdoi2010]大陸爭霸