Leetcode815. Bus Routes

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We have a list of bus routes. Each routes[i] is a bus route that the i-th bus repeats forever. For example if routes[0] = [1, 5, 7], this means that the first bus (0-th indexed) travels in the sequence 1->5->7->1->5->7->1->… forever.

We start at bus stop S (initially not on a bus), and we want to go to bus stop T. Travelling by buses only, what is the least number of buses we must take to reach our destination? Return -1 if it is not possible.

Example:

Input:

routes = [[1, 2, 7], [3, 6, 7]]

S = 1

T = 6

Output: 2

Explanation:

The best strategy is take the first bus to the bus stop 7, then take the second bus to the bus stop 6.

Note:

1 <= routes.length <= 500.
1 <= routes[i].length <= 500.
0 <= routes[i][j] < 10 ^ 6.

思路

queue存储站点 如果站点是同一辆公交车可达就跳过（剪枝)

要不就把该站点加入到队列中

代码

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class Solution {
public:
    int numBusesToDestination(vector<vector<int>>& routes, int S, int T) {
    queue<int> q;
    
    map<int,vector<int> > rmap;
        for(int i=0;i<routes.size();i++){
            for(int j=0;j<routes[i].size();j++){
               rmap[routes[i][j]].push_back(i);
                        
            }  
        }
        vector<int> vis(500,0);
        q.push(S);
        if(S==T) return 0;
        int steps=0;
        while(!q.empty()){
                int size=q.size();
            while(size--)
         {
            int t=q.front();
            q.pop();
            for(auto bus:rmap[t]){
                if(vis[bus]) continue;
                vis[bus]=1;
                for(auto stop:routes[bus]){
                    if(T==stop) return steps+1;
                    q.push(stop);         
                }
                
                
            }
                              
       }
        
            steps++;
            
        }
        return -1;
        
    }
};