Leetcode 815. Bus Routes

Problem:

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.

 

Solution:

　　这道题看到最少数量的公交车就应该有使用BFS的预感，先来说说我最开始考虑的解法，首先用一个哈希表记录每一个站点stop经过的所有线路的公交车，公交车编号为routes的索引。然后用visited记录访问过的公交车编号。BFS肯定是要维护一个队列的，里面的元素pair第一个元素代表第first路公交，第二个元素代表该公交经过第second站。首先将所有包含S站的公交全部放入队列，表示只需要一量公交可以到达的所有站点，然后在循环中记录下队列长度，然后将所有未访问过的公交车可能到达的所有站点加入队列中，同时pop开始元素，直到找到T为止，理论上说这个解法的时间复杂度是没问题的，可能是由于测试用例的缘故时间稍慢。

class Solution {
public:
    int numBusesToDestination(vector<vector<int>>& routes, int S, int T) {
        if(S == T) return 0;
        unordered_map<int,vector<int>> um;
        for(int i = 0;i != routes.size();++i){
            for(int j = 0;j != routes[i].size();++j){
                um[routes[i][j]].push_back(i);
            }
        }
        vector<int> visited(routes.size(),false);
        queue<pair<int,int>> q;
        for(int i = 0;i != um[S].size();++i){
            int busline = um[S][i];
            for(int j = 0;j != routes[busline].size();++j){
                q.push(make_pair(busline,routes[busline][j]));
                if(routes[busline][j] == T)
                    return 1;
            }
            visited[busline] = true;
        }
        int result = 2;
        while(!q.empty()){
            int m = q.size();
            for(int i = 0;i != m;++i){
                int busline = q.front().first;
                int stop = q.front().second;
                q.pop();
                for(int k = 0;k != um[stop].size();++k){
                    if(visited[um[stop][k]]) continue;
                    for(int t = 0;t != routes[um[stop][k]].size();++t){
                        q.push(make_pair(um[stop][k],routes[um[stop][k]][t]));
                        if(routes[um[stop][k]][t] == T) return result;
                    }
                    visited[um[stop][k]] = true;
                }
            }
            result++;
        }
        return -1;
    }
};

　　接下来讲另一种解法，大致思路类似，不同的是这回pair中第一个元素记录站点，第二个元素记录坐过的公交车数量，visited记录做过的站点，然后在队列中不断添加该公交车可以换乘到达的站点。注意为什么push的一定是可以换乘到达的站点，这是由于同一条线路上的其他站点必然已经添加过队列了（细细体会这个细节）。

Code:

 

class Solution {
public:
    int numBusesToDestination(vector<vector<int>>& routes, int S, int T) {
        if(S == T) return 0;
        unordered_map<int,vector<int>> um;
        for(int i = 0;i != routes.size();++i){
            for(int j = 0;j != routes[i].size();++j){
                um[routes[i][j]].push_back(i);
            }
        }
        unordered_set<int> visited;
        visited.insert(S);
        queue<pair<int,int>> q;
        q.push(make_pair(S,0));
        while(!q.empty()){
            int stop = q.front().first;
            int transfer = q.front().second;
            q.pop();
            for(int i = 0;i != um[stop].size();++i){
                for(int j = 0;j != routes[um[stop][i]].size();++j){
                    if(T == routes[um[stop][i]][j]) return transfer+1;
                    if(visited.find(routes[um[stop][i]][j]) == visited.end()){
                        visited.insert(routes[um[stop][i]][j]);
                        q.push(make_pair(routes[um[stop][i]][j],transfer+1));
                    }
                }
                routes[um[stop][i]].clear();
            }
        }
        return -1;
    }
};