题面
Sol
走到一个点的前提是所有的影响它的点都走到，并且它的时间为那些点的时间\(max\)与自己的最短路的\(max\)
考虑\(Dijkstra\)
每次松弛时候，把它影响到的点的防护罩减\(1\)
如果某个点没有后继影响它的节点就丢到大根堆内
然后这个继续松弛
# 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;
}