【每日算法Day 62】LeetCode 815. 公交路线

题目描述

示例1

输入：
routes = [[1, 2, 7], [3, 6, 7]]
S = 1
T = 6
输出：
2
解释：
最优策略是先乘坐第一辆公交车到达车站 7, 然后换乘第二辆公交车到车站 6。

提示

题解

代码

c++

class Solution {
public:
    int numBusesToDestination(vector<vector<int>>& routes, int S, int T) {
        if (S == T) return 0;
        int n = routes.size();
        for (int i = 0; i < n; ++i) {
            sort(routes[i].begin(), routes[i].end());
            routes[i].erase(unique(routes[i].begin(), routes[i].end()), routes[i].end());
        }
        routes.push_back({S});
        routes.push_back({T});
        vector<vector<int>> G = buildGraph(routes, S, T);
        return BFS(G);
    }

    vector<vector<int>> buildGraph(vector<vector<int>>& routes, int S, int T) {
        int n = routes.size();
        vector<vector<int>> G(n);
        for (int i = 0; i < n; ++i) {
            for (int j = i+1; j < n; ++j) {
                int su = routes[i].size(), sv = routes[j].size();
                int u = 0, v = 0;
                while (u < su && v < sv) {
                    if (routes[i][u] < routes[j][v]) ++u;
                    else if (routes[i][u] > routes[j][v]) ++v;
                    else {
                        G[i].push_back(j);
                        G[j].push_back(i);
                        break;
                    }
                }
            }
        }
        return G;
    }

    int BFS(vector<vector<int>>& G) {
        int n = G.size();
        int S = n - 2;
        int T = n - 1;
        vector<int> dis(n, -1);
        queue<int> Q;
        Q.push(S);
        dis[S] = 0;
        while (!Q.empty()) {
            int u = Q.front();
            Q.pop();
            for (int i = 0, sz = G[u].size(); i < sz; ++i) {
                int v = G[u][i];
                if (dis[v] == -1) {
                    Q.push(v);
                    dis[v] = dis[u] + 1;
                    if (v == T) return dis[v]-1;
                }
            }
        }
        return -1;
    }
};