【Leetcode】743. Network Delay Time

There are N network nodes, labelled 1 to N.

Given times, a list of travel times as directed edges times[i] = (u, v, w), where u is the source node, v is the target node, and w is the time it takes for a signal to travel from source to target.

Now, we send a signal from a certain node K. How long will it take for all nodes to receive the signal? If it is impossible, return -1.

Note:

1. N will be in the range [1, 100].
2. K will be in the range [1, N].
3. The length of times will be in the range [1, 6000].
4. All edges times[i] = (u, v, w) will have 1 <= u, v <= N and 1 <= w <= 100.

题解：有向图中求起点所有最短路径中最长的一条，直接用DIjkstra算法即可，本题刚开始需要理解清楚题意，不是求起点到所有节点的路径总和，因为信息是同时发送的所需时间取决于最长的路径。注意下标是从1开始所以用for-each不能用

class Solution {
public:
    int networkDelayTime(vector<vector<int>>& times, int N, int K) {
        vector<int> d(N+1,0);
        vector<bool> vis(N+1,false);
        const int oo = 1000;
        vector<vector<int>> G(N+1,vector<int>(N+1,oo));
        for(int i=0;i<times.size();i++){
            int u=times[i][0],v=times[i][1],w=times[i][2];
            G[u][v]=w;
        }
        
        Dijkstra(K,G,d,vis);
        int ans=0;
        for(int i=1;i<d.size();i++)
        {
            if(d[i]==oo)return -1;
            ans=max(ans,d[i]);
        }
        return ans;
    }

    void Dijkstra(int s,vector<vector<int>> G,vector<int>& d,vector<bool> vis)
    {
        const int oo = 1000;
        fill(d.begin(),d.end(),oo);
        d[s]=0;
        for(int i=1;i<G.size();i++)
        {
            int u=-1,MIN=oo;
            for(int j=1;j<G.size();j++){
                if(vis[j]==false&&d[j]<MIN){
                    u=j;
                    MIN=d[j];
                }
            }
            if(u==-1)return;
            vis[u]=true;
            for(int v=1;v<G.size();v++){
                if(vis[v]==false&&G[u][v]!=oo&&d[u]+G[u][v]<d[v]){
                    d[v]=d[u]+G[u][v];
                }
            }
        }
    }
};