題目連結：https://www.lydsy.com/JudgeOnline/problem.php?id=1922

漲姿勢了，，，這下不只是複習了。。。

此題可以認為是一種變相的最短路，嗯，有限制條件的最短路。

設dist1[i]表示到達某個結點的最短時間，dist2[i]表示可以進入某個結點的最短時間，顯然，真正進入的最短時間是max(dist1[i],dist2[i])。對於每一個用來更新其他結點的結點，我們用他實際進入的時間去更新與他相鄰的城市的到達時間，依賴於他的城市的可進入時間，並且依賴於他的城市的入度減1。每次更新都要把入度為0的點加入優先佇列，不管是原先就為0，還是減1後為0。

另外提一點，不要把讀入優化直接傳入引數，順序會變！別問我怎麼知道的。。。

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 #include <queue>
 5  
 6 using namespace std;
 7  
 8 inline int get_num() {
 9     int num = 0;
10     char c = getchar();
11     while (c < '0'
 
 || c > '9') c = getchar();
12     while (c >= '0' && c <= '9')
13         num = num * 10 + c - '0', c = getchar();
14     return num;
15 }
16  
17 const int maxn = 3e3 + 5, maxm = 7e4 + 5, inf = 0x3f3f3f3f;
18  
19 int head[maxn], eid, head2[maxn], eid2, ind[maxn];
20  
21 struct
 
 Edge {
22     int v, w, next;
23 } edge[maxm], edge2[maxm];
24  
25 inline void insert(int u, int v, int w) {
26     edge[++eid].v = v;
27     edge[eid].w = w;
28     edge[eid].next = head[u];
29     head[u] = eid;
30 }
31  
32 inline void insert2(int u, int v) {
33     edge2[++eid2].v = v;
34     edge2[eid2].next = head2[u];
35     head2[u] = eid2;
36 }
37  
38 int dist1[maxn], dist2[maxn], vis[maxn];
39  
40 struct node {
41     int id, dist;
42     node(int id, int dist) : id(id), dist(dist) {}
43     bool operator < (const node& rhs) const {
44         return dist > rhs.dist;
45     }
46 };
47  
48 priority_queue<node> q;
49  
50 inline void dijkstra() {
51     memset(dist1, inf, sizeof(dist1));
52     dist1[1] = 0;
53     q.push(node(1, 0));
54     while (!q.empty()) {
55         int u = q.top().id;
56         q.pop();
57         if (vis[u]) continue;
58         vis[u] = 1;
59         int td = max(dist1[u], dist2[u]);
60         for (int p = head[u]; p; p = edge[p].next) {
61             int v = edge[p].v, w = edge[p].w;
62             if (dist1[v] > td + w) {
63                 dist1[v] = td + w;
64                 if (!ind[v]) q.push(node(v, max(dist1[v], dist2[v])));
65             }
66         }
67         for (int p = head2[u]; p; p = edge2[p].next) {
68             int v = edge2[p].v;
69             --ind[v], dist2[v] = max(dist2[v], td);
70             if (!ind[v]) q.push(node(v, max(dist1[v], dist2[v])));
71         }
72     }
73 }
74  
75 int main() {
76     int n = get_num(), m = get_num();
77     for (int i = 1; i <= m; ++i) {
78         int u = get_num(), v = get_num(), w = get_num();
79         if (u != v) insert(u, v, w);
80     }
81     for (int i = 1; i <= n; ++i) {
82         ind[i] = get_num();
83         for (int j = 1; j <= ind[i]; ++j) {
84             int x = get_num();
85             insert2(x, i);
86         }
87     }
88     dijkstra();
89     printf("%d", max(dist1[n], dist2[n]));
90     return 0;
91 }