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41 changes: 41 additions & 0 deletions
...estination-in-Directed-Graph/3604.Minimum-Time-to-Reach-Destination-in-Directed-Graph.cpp
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struct state { | ||
int node, time; | ||
bool operator<(state const& o) const { | ||
return time > o.time; | ||
} | ||
}; | ||
class Solution { | ||
vector<vector<int>>next[100005]; | ||
public: | ||
int minTime(int n, vector<vector<int>>& edges) { | ||
for (auto& e:edges) { | ||
int u=e[0], v=e[1], start=e[2], end=e[3]; | ||
next[u].push_back({v, start, end}); | ||
} | ||
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vector<int>visited(n, -1); | ||
priority_queue<state>pq; | ||
pq.push({0,0}); | ||
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while (!pq.empty()) { | ||
auto [node, time] = pq.top(); | ||
// cout<<node<<" "<<time<<endl; | ||
pq.pop(); | ||
if (visited[node]!=-1) continue; | ||
visited[node] = time; | ||
if (node==n-1) return time; | ||
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for (auto nxt: next[node]) { | ||
int v = nxt[0], s = nxt[1], e = nxt[2]; | ||
if (time < s && (visited[v]==-1 || visited[v]>s+1)) { | ||
pq.push({v, s+1}); | ||
} | ||
if (time >= s && time <=e && (visited[v]==-1 || visited[v]>time+1)) { | ||
pq.push({v, time+1}); | ||
} | ||
} | ||
} | ||
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return -1; | ||
} | ||
}; |
12 changes: 12 additions & 0 deletions
BFS/3604.Minimum-Time-to-Reach-Destination-in-Directed-Graph/Readme.md
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### 3604.Minimum-Time-to-Reach-Destination-in-Directed-Graph | ||
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常规的Dijkstra的模版题。本题优先队里的元素需要定义状态 | ||
```cpp | ||
struct state { | ||
int node, time; | ||
bool operator<(state const& o) const { | ||
return time > o.time; | ||
} | ||
}; | ||
``` | ||
对于出队列的一个状态{node,time},我们可知到达node的最短时间就是time。然后我们检查它周围的路径{v,start,end}。如果time<start,那么我们就把{v,start+1}加入优先队列。如果time在[start,end]时段内,那么我们就把{v,time+1}加入优先队列。 |
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