Breadth-first Search - Hard - Question 1

Breadth-first Search - Hard - Question 1

815. Bus Routes

You are given an array routes representing bus routes where routes[i] is a bus route that the ith bus repeats forever.

For example, if routes[0] = [1, 5, 7], this means that the 0th bus travels in the sequence 1 -> 5 -> 7 -> 1 -> 5 -> 7 -> 1 -> ... forever.

You will start at the bus stop source (You are not on any bus initially), and you want to go to the bus stop target. You can travel between bus stops by buses only.

Return the least number of buses you must take to travel from source to target. Return -1 if it is not possible.

Constraints:

1 <= routes.length <= 500.

1 <= routes[i].length <= 10^5

All the values of routes[i] are unique.

sum(routes[i].length) <= 10^5

0 <= routes[i][j] < 10^6

0 <= source, target < 10^6

Analysis:

One of key steps to this question is to build the connections.

For example, once we start with the source stop, then we know what is the next stops to go. With the connections, we should quickly know what they are (without scanning the raw data).

This connection is usually called a graph.

After having this graph, we can do a regular breadth-first search (bfs) for the shortest distance. And also, to avoid redundant visits, we can use a visit array to record this.

See the code below:

class Solution {
public:
    int numBusesToDestination(vector<vector<int>>& routes, int source, int target) {
        int res = 0, n = routes.size();
        if(source == target) return res;
        unordered_map<int, vector<int>> data;
        for(int i=0; i<n; ++i) {
            for(auto &a : routes[i]) {
                data[a].push_back(i);
            }
        }
        vector<int> hs(n, 0), vt(n, 0);
        for(auto &a : data[target]) hs[a] = 1;
        queue<int> q;
        for(auto &a : data[source]) {
            q.push(a);
            vt[a] = 1;
        }
        while(q.size()) {
            ++res;
            int k = q.size();
            for(int i=0; i<k; ++i) {
                auto t = q.front();
                q.pop();
                if(hs[t]) return res;
                for(auto &b : routes[t]) {
                    for(auto &a : data[b]) {
                        if(!vt[a]) {
                            q.push(a);
                            vt[a] = 1;
                        }
                    }
                }
            }
        }
        return -1;
    }
};

Upper Layer