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847. Shortest Path Visiting All Nodes
Description
You have an undirected, connected graph of n nodes labeled from 0 to n - 1. You are given an array graph where graph[i] is a list of all the nodes connected with node i by an edge.
Return the length of the shortest path that visits every node. You may start and stop at any node, you may revisit nodes multiple times, and you may reuse edges.
Example 1:
Input: graph = [[1,2,3],[0],[0],[0]] Output: 4 Explanation: One possible path is [1,0,2,0,3]
Example 2:
Input: graph = [[1],[0,2,4],[1,3,4],[2],[1,2]] Output: 4 Explanation: One possible path is [0,1,4,2,3]
Constraints:
n == graph.length1 <= n <= 120 <= graph[i].length < ngraph[i]does not containi.- If
graph[a]containsb, thengraph[b]containsa. - The input graph is always connected.
Solutions
Solution 1
Because each edge has the same weight, the shortest path can be solution by using BFS, and the access of the point can be recorded by state compression.
Solution 2
Because the weight of each edge is the same, the shortest path can be obtained using BFS. In the process, state compression can be used to record the access status of the node. In addition, the same node u and the corresponding node access situation need to be guaranteed to be searched only once, so vis(u, state) can be used to indicate whether it has been searched to prevent invalid repeated searches.
This question also belongs to the BFS minimum step model, and the A* algorithm can be used to optimize the search.
The main idea of A* algorithm is as follows:
- Convert the BFS queue to a priority queue (small root heap);
- Each element in the queue is
(dist[state] + f(state), state),dist[state]represents the distance from the starting point to the current state, andf(state)represents the estimated distance from the current state to the end point. The sum of these two distances is used as the basis for heap sorting; - When the end point is dequeued for the first time, it means that the shortest path from the starting point to the end point is found and the corresponding step is returned directly;
f(state)is the valuation function, and the valuation function must satisfyf(state) <= g(state), whereg(state)represents the true distance from state to the end point;- The A* algorithm can only guarantee that when the end point is dequeued for the first time, it will find a minimum path from the starting point to the end point. It cannot guarantee that when other points are dequeued, it will also be the shortest path from the starting point to the current point.
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class Solution { public int shortestPathLength(int[][] graph) { int n = graph.length; Deque<int[]> q = new ArrayDeque<>(); boolean[][] vis = new boolean[n][1 << n]; for (int i = 0; i < n; ++i) { q.offer(new int[] {i, 1 << i}); vis[i][1 << i] = true; } for (int ans = 0;; ++ans) { for (int k = q.size(); k > 0; --k) { var p = q.poll(); int i = p[0], st = p[1]; if (st == (1 << n) - 1) { return ans; } for (int j : graph[i]) { int nst = st | 1 << j; if (!vis[j][nst]) { vis[j][nst] = true; q.offer(new int[] {j, nst}); } } } } } } // Solution 2 class Solution { private int n; public int shortestPathLength(int[][] graph) { n = graph.length; int[][] dist = new int[n][1 << n]; for (int i = 0; i < n; ++i) { Arrays.fill(dist[i], Integer.MAX_VALUE); } PriorityQueue<int[]> q = new PriorityQueue<>(Comparator.comparingInt(a -> a[0])); for (int i = 0; i < n; ++i) { q.offer(new int[] {f(1 << i), i, 1 << i}); dist[i][1 << i] = 0; } while (!q.isEmpty()) { int[] p = q.poll(); int u = p[1], state = p[2]; if (state == (1 << n) - 1) { return dist[u][state]; } for (int v : graph[u]) { int nxt = state | (1 << v); if (dist[v][nxt] > dist[u][state] + 1) { dist[v][nxt] = dist[u][state] + 1; q.offer(new int[] {dist[v][nxt] + f(nxt), v, nxt}); } } } return 0; } private int f(int state) { int ans = 0; for (int i = 0; i < n; ++i) { if (((state >> i) & 1) == 0) { ++ans; } } return ans; } } -
class Solution { public: int shortestPathLength(vector<vector<int>>& graph) { int n = graph.size(); queue<pair<int, int>> q; bool vis[n][1 << n]; memset(vis, false, sizeof(vis)); for (int i = 0; i < n; ++i) { q.emplace(i, 1 << i); vis[i][1 << i] = true; } for (int ans = 0;; ++ans) { for (int k = q.size(); k; --k) { auto [i, st] = q.front(); q.pop(); if (st == (1 << n) - 1) { return ans; } for (int j : graph[i]) { int nst = st | 1 << j; if (!vis[j][nst]) { vis[j][nst] = true; q.emplace(j, nst); } } } } } }; // Solution 2 class Solution { public: int n; int shortestPathLength(vector<vector<int>>& graph) { n = graph.size(); priority_queue<tuple<int, int, int>, vector<tuple<int, int, int>>, greater<tuple<int, int, int>>> q; vector<vector<int>> dist(n, vector<int>(1 << n, INT_MAX)); for (int i = 0; i < n; ++i) { q.push({f(1 << i), i, 1 << i}); dist[i][1 << i] = 0; } while (!q.empty()) { auto [_, u, state] = q.top(); q.pop(); if (state == (1 << n) - 1) return dist[u][state]; for (int v : graph[u]) { int nxt = state | (1 << v); if (dist[v][nxt] > dist[u][state] + 1) { dist[v][nxt] = dist[u][state] + 1; q.push({dist[v][nxt] + f(nxt), v, nxt}); } } } return 0; } int f(int state) { int ans = 0; for (int i = 0; i < n; ++i) if (((state >> i) & 1) == 0) ++ans; return ans; } }; -
class Solution: def shortestPathLength(self, graph: List[List[int]]) -> int: n = len(graph) q = deque() vis = set() for i in range(n): q.append((i, 1 << i)) vis.add((i, 1 << i)) ans = 0 while 1: for _ in range(len(q)): i, st = q.popleft() if st == (1 << n) - 1: return ans for j in graph[i]: nst = st | 1 << j if (j, nst) not in vis: vis.add((j, nst)) q.append((j, nst)) ans += 1 # Solution 2 class Solution: def shortestPathLength(self, graph: List[List[int]]) -> int: n = len(graph) def f(state): return sum(((state >> i) & 1) == 0 for i in range(n)) q = [] dist = [[inf] * (1 << n) for _ in range(n)] for i in range(n): heappush(q, (f(1 << i), i, 1 << i)) dist[i][1 << i] = 0 while q: _, u, state = heappop(q) if state == (1 << n) - 1: return dist[u][state] for v in graph[u]: nxt = state | (1 << v) if dist[v][nxt] > dist[u][state] + 1: dist[v][nxt] = dist[u][state] + 1 heappush(q, (dist[v][nxt] + f(nxt), v, nxt)) return 0 -
func shortestPathLength(graph [][]int) int { n := len(graph) q := [][2]int{} vis := make([][]bool, n) for i := range vis { vis[i] = make([]bool, 1<<n) vis[i][1<<i] = true q = append(q, [2]int{i, 1 << i}) } for ans := 0; ; ans++ { for k := len(q); k > 0; k-- { p := q[0] q = q[1:] i, st := p[0], p[1] if st == (1<<n)-1 { return ans } for _, j := range graph[i] { nst := st | 1<<j if !vis[j][nst] { vis[j][nst] = true q = append(q, [2]int{j, nst}) } } } } } -
function shortestPathLength(graph: number[][]): number { const n = graph.length; const q: number[][] = []; const vis: boolean[][] = new Array(n).fill(false).map(() => new Array(1 << n).fill(false)); for (let i = 0; i < n; ++i) { q.push([i, 1 << i]); vis[i][1 << i] = true; } for (let ans = 0; ; ++ans) { for (let k = q.length; k; --k) { const [i, st] = q.shift()!; if (st === (1 << n) - 1) { return ans; } for (const j of graph[i]) { const nst = st | (1 << j); if (!vis[j][nst]) { vis[j][nst] = true; q.push([j, nst]); } } } } } -
use std::collections::VecDeque; impl Solution { #[allow(dead_code)] pub fn shortest_path_length(graph: Vec<Vec<i32>>) -> i32 { let n = graph.len(); let mut vis = vec![vec![false; 1 << n]; n]; let mut q = VecDeque::new(); // Initialize the queue for i in 0..n { q.push_back(((i, 1 << i), 0)); vis[i][1 << i] = true; } // Begin BFS while !q.is_empty() { let ((i, st), count) = q.pop_front().unwrap(); if st == (1 << n) - 1 { return count; } // If the path has not been visited for j in &graph[i] { let nst = st | (1 << *j); if !vis[*j as usize][nst] { q.push_back(((*j as usize, nst), count + 1)); vis[*j as usize][nst] = true; } } } -1 } }