Created
May 14, 2018 14:26
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class Solution { | |
public: | |
int networkDelayTime(vector<vector<int>>& times, int N, int K) { | |
// first v, second w | |
vector<vector<pair<int, int> > > adj(N+1, vector<pair<int, int> >()); | |
// dist and parent | |
vector<pair<int, int> > dist_p(N+1, make_pair(INT_MAX, -1)); | |
// visited | |
vector<bool > visited(N+1, false); | |
// initialize | |
for (vector<vector<int> >::iterator i = times.begin(); i != times.end(); ++i) | |
{ | |
adj[(*i)[0]].push_back(make_pair((*i)[1], (*i)[2])); | |
} | |
dist_p[K] = make_pair(0, K); | |
// dijkstra | |
for (int i = 1; i < N; ++i) | |
{ | |
// find the min dist node | |
int u = K; | |
int m_dist = INT_MAX; | |
for (int i = 0; i < dist_p.size(); ++i) | |
{ | |
if(dist_p[i].first < m_dist && !visited[i]){ | |
u = i; | |
m_dist = dist_p[i].first; | |
} | |
} | |
visited[u] = true; | |
//relax | |
for (int i = 0; i < adj[u].size(); ++i) | |
{ | |
int v = adj[u][i].first; | |
int distuv = adj[u][i].second; | |
if(dist_p[v].first > distuv + dist_p[u].first){ | |
dist_p[v].first = distuv + dist_p[u].first; | |
dist_p[v].second = u; | |
} | |
} | |
} | |
// test whether all the node has path to K | |
int far = 0; | |
for (int i = 1; i < N+1; ++i) | |
{ | |
if(dist_p[i].first > far){ | |
far = dist_p[i].first; | |
} | |
} | |
if(INT_MAX == far){ | |
return -1; | |
}else{ | |
return far; | |
} | |
} | |
}; |
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