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743. 网络延迟时间

English Version

题目描述

有 n 个网络节点，标记为 1 到 n。

给你一个列表 times，表示信号经过 有向 边的传递时间。 times[i] = (ui, vi, wi)，其中 ui 是源节点，vi 是目标节点， wi 是一个信号从源节点传递到目标节点的时间。

现在，从某个节点 K 发出一个信号。需要多久才能使所有节点都收到信号？如果不能使所有节点收到信号，返回 -1 。

 

示例 1：

输入：times = [[2,1,1],[2,3,1],[3,4,1]], n = 4, k = 2
输出：2

示例 2：

输入：times = [[1,2,1]], n = 2, k = 1
输出：1

示例 3：

输入：times = [[1,2,1]], n = 2, k = 2
输出：-1

 

提示：

  • 1 <= k <= n <= 100
  • 1 <= times.length <= 6000
  • times[i].length == 3
  • 1 <= ui, vi <= n
  • ui != vi
  • 0 <= wi <= 100
  • 所有 (ui, vi) 对都 互不相同（即，不含重复边）

解法

设 n 表示点数，m 表示边数。

方法一：朴素 Dijkstra 算法

时间复杂度 O(n²+m)。

方法二：堆优化 Dijkstra 算法

时间复杂度 O(mlogn)。

方法三：Bellman Ford 算法

时间复杂度 O(nm)。

方法四：SPFA 算法

时间复杂度，平均情况下 O(m)，最坏情况下 O(nm)。

Python3

朴素 Dijkstra 算法：

class Solution:
    def networkDelayTime(self, times: List[List[int]], n: int, k: int) -> int:
        INF = 0x3f3f
        dist = [INF] * n
        vis = [False] * n
        g = [[INF] * n for _ in range(n)]
        for u, v, w in times:
            g[u - 1][v - 1] = w
        dist[k - 1] = 0
        for _ in range(n):
            t = -1
            for j in range(n):
                if not vis[j] and (t == -1 or dist[t] > dist[j]):
                    t = j
            vis[t] = True
            for j in range(n):
                dist[j] = min(dist[j], dist[t] + g[t][j])
        ans = max(dist)
        return -1 if ans == INF else ans

堆优化 Dijkstra 算法：

class Solution:
    def networkDelayTime(self, times: List[List[int]], n: int, k: int) -> int:
        INF = 0x3f3f
        g = defaultdict(list)
        for u, v, w in times:
            g[u - 1].append((v - 1, w))
        dist = [INF] * n
        dist[k - 1] = 0
        q = [(0, k - 1)]
        while q:
            _, u = heappop(q)
            for v, w in g[u]:
                if dist[v] > dist[u] + w:
                    dist[v] = dist[u] + w
                    heappush(q, (dist[v], v))
        ans = max(dist)
        return -1 if ans == INF else ans

Bellman Ford 算法：

class Solution:
    def networkDelayTime(self, times: List[List[int]], n: int, k: int) -> int:
        INF = 0x3f3f
        dist = [INF] * n
        dist[k - 1] = 0
        for _ in range(n):
            backup = dist[:]
            for u, v, w in times:
                dist[v - 1] = min(dist[v - 1], dist[u - 1] + w)
        ans = max(dist)
        return -1 if ans == INF else ans

SPFA 算法：

class Solution:
    def networkDelayTime(self, times: List[List[int]], n: int, k: int) -> int:
        INF = 0x3f3f
        dist = [INF] * n
        vis = [False] * n
        g = defaultdict(list)
        for u, v, w in times:
            g[u - 1].append((v - 1, w))
        k -= 1
        dist[k] = 0
        q = deque([k])
        vis[k] = True
        while q:
            u = q.popleft()
            vis[u] = False
            for v, w in g[u]:
                if dist[v] > dist[u] + w:
                    dist[v] = dist[u] + w
                    if not vis[v]:
                        q.append(v)
                        vis[v] = True
        ans = max(dist)
        return -1 if ans == INF else ans

Java

朴素 Dijkstra 算法：

class Solution {
    private static final int INF = 0x3f3f;

    public int networkDelayTime(int[][] times, int n, int k) {
        int[][] g = new int[n][n];
        int[] dist = new int[n];
        boolean[] vis = new boolean[n];
        for (int i = 0; i < n; ++i) {
            dist[i] = INF;
            Arrays.fill(g[i], INF);
        }
        for (int[] t : times) {
            g[t[0] - 1][t[1] - 1] = t[2];
        }
        dist[k - 1] = 0;
        for (int i = 0; i < n; ++i) {
            int t = -1;
            for (int j = 0; j < n; ++j) {
                if (!vis[j] && (t == -1 || dist[t] > dist[j])) {
                    t = j;
                }
            }
            vis[t] = true;
            for (int j = 0; j < n; ++j) {
                dist[j] = Math.min(dist[j], dist[t] + g[t][j]);
            }
        }
        int ans = 0;
        for (int d : dist) {
            ans = Math.max(ans, d);
        }
        return ans == INF ? -1 : ans;
    }
}

堆优化 Dijkstra 算法：

class Solution {
    private static final int INF = 0x3f3f;

    public int networkDelayTime(int[][] times, int n, int k) {
        List<int[]>[] g = new List[n];
        int[] dist = new int[n];
        for (int i = 0; i < n; ++i) {
            dist[i] = INF;
            g[i] = new ArrayList<>();
        }
        for (int[] t : times) {
            g[t[0] - 1].add(new int[]{t[1] - 1, t[2]});
        }
        dist[k - 1] = 0;
        PriorityQueue<int[]> q = new PriorityQueue<>(Comparator.comparingInt(a -> a[0]));
        q.offer(new int[]{0, k - 1});
        while (!q.isEmpty()) {
            int[] p = q.poll();
            int u = p[1];
            for (int[] ne : g[u]) {
                int v = ne[0], w = ne[1];
                if (dist[v] > dist[u] + w) {
                    dist[v] = dist[u] + w;
                    q.offer(new int[]{dist[v], v});
                }
            }
        }
        int ans = 0;
        for (int d : dist) {
            ans = Math.max(ans, d);
        }
        return ans == INF ? -1 : ans;
    }
}

Bellman Ford 算法：

class Solution {
    private static final int INF = 0x3f3f;

    public int networkDelayTime(int[][] times, int n, int k) {
        int[] dist = new int[n];
        int[] backup = new int[n];
        Arrays.fill(dist, INF);
        dist[k - 1] = 0;
        for (int i = 0; i < n; ++i) {
            System.arraycopy(dist, 0, backup, 0, n);
            for (int[] t : times) {
                int u = t[0] - 1, v = t[1] - 1, w = t[2];
                dist[v] = Math.min(dist[v], backup[u] + w);
            }
        }
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            ans = Math.max(ans, dist[i]);
        }
        return ans == INF ? -1 : ans;
    }
}

SPFA 算法：

class Solution {
    private static final int INF = 0x3f3f;

    public int networkDelayTime(int[][] times, int n, int k) {
        int[] dist = new int[n];
        boolean[] vis = new boolean[n];
        List<int[]>[] g = new List[n];
        for (int i = 0; i < n; ++i) {
            dist[i] = INF;
            g[i] = new ArrayList<>();
        }
        for (int[] t : times) {
            int u = t[0] - 1, v = t[1] - 1, w = t[2];
            g[u].add(new int[]{v, w});
        }
        --k;
        dist[k] = 0;
        Deque<Integer> q = new ArrayDeque<>();
        q.offer(k);
        vis[k] = true;
        while (!q.isEmpty()) {
            int u = q.poll();
            vis[u] = false;
            for (int[] ne : g[u]) {
                int v = ne[0], w = ne[1];
                if (dist[v] > dist[u] + w) {
                    dist[v] = dist[u] + w;
                    if (!vis[v]) {
                        q.offer(v);
                        vis[v] = true;
                    }
                }
            }
        }
        int ans = 0;
        for (int i = 0; i < n; ++i) {
            ans = Math.max(ans, dist[i]);
        }
        return ans == INF ? -1 : ans;
    }
}

Go

朴素 Dijkstra 算法：

func networkDelayTime(times [][]int, n int, k int) int {
	const inf = 0x3f3f
	dist := make([]int, n)
	vis := make([]bool, n)
	g := make([][]int, n)
	for i := range dist {
		dist[i] = inf
		g[i] = make([]int, n)
		for j := range g[i] {
			g[i][j] = inf
		}
	}
	for _, t := range times {
		g[t[0]-1][t[1]-1] = t[2]
	}
	dist[k-1] = 0
	for i := 0; i < n; i++ {
		t := -1
		for j := 0; j < n; j++ {
			if !vis[j] && (t == -1 || dist[t] > dist[j]) {
				t = j
			}
		}
		vis[t] = true
		for j := 0; j < n; j++ {
			dist[j] = min(dist[j], dist[t]+g[t][j])
		}
	}
	ans := 0
	for _, v := range dist {
		ans = max(ans, v)
	}
	if ans == inf {
		return -1
	}
	return ans
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

堆优化 Dijkstra 算法：

const Inf = 0x3f3f3f3f

type pair struct {
	first  int
	second int
}

var _ heap.Interface = (*pairs)(nil)

type pairs []pair

func (a pairs) Len() int { return len(a) }
func (a pairs) Less(i int, j int) bool {
	return a[i].first < a[j].first || a[i].first == a[j].first && a[i].second < a[j].second
}
func (a pairs) Swap(i int, j int)   { a[i], a[j] = a[j], a[i] }
func (a *pairs) Push(x interface{}) { *a = append(*a, x.(pair)) }
func (a *pairs) Pop() interface{}   { l := len(*a); t := (*a)[l-1]; *a = (*a)[:l-1]; return t }

func networkDelayTime(times [][]int, n int, k int) int {
	graph := make([]pairs, n)
	for _, time := range times {
		from, to, time := time[0]-1, time[1]-1, time[2]
		graph[from] = append(graph[from], pair{to, time})
	}

	dis := make([]int, n)
	for i := range dis {
		dis[i] = Inf
	}
	dis[k-1] = 0

	vis := make([]bool, n)
	h := make(pairs, 0)
	heap.Push(&h, pair{0, k - 1})
	for len(h) > 0 {
		from := heap.Pop(&h).(pair).second
		if vis[from] {
			continue
		}
		vis[from] = true
		for _, e := range graph[from] {
			to, d := e.first, dis[from]+e.second
			if d < dis[to] {
				dis[to] = d
				heap.Push(&h, pair{d, to})
			}
		}
	}

	ans := math.MinInt32
	for _, d := range dis {
		ans = max(ans, d)
	}
	if ans == Inf {
		return -1
	}
	return ans
}

func max(x, y int) int {
	if x > y {
		return x
	}
	return y
}

Bellman Ford 算法：

func networkDelayTime(times [][]int, n int, k int) int {
	const inf = 0x3f3f
	dist := make([]int, n)
	backup := make([]int, n)
	for i := range dist {
		dist[i] = inf
	}
	dist[k-1] = 0
	for i := 0; i < n; i++ {
		copy(backup, dist)
		for _, e := range times {
			u, v, w := e[0]-1, e[1]-1, e[2]
			dist[v] = min(dist[v], backup[u]+w)
		}
	}
	ans := 0
	for _, v := range dist {
		ans = max(ans, v)
	}
	if ans == inf {
		return -1
	}
	return ans
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

SPFA 算法：

func networkDelayTime(times [][]int, n int, k int) int {
	const inf = 0x3f3f
	dist := make([]int, n)
	vis := make([]bool, n)
	g := make([][][]int, n)
	for i := range dist {
		dist[i] = inf
	}
	for _, t := range times {
		u, v, w := t[0]-1, t[1]-1, t[2]
		g[u] = append(g[u], []int{v, w})
	}
	k--
	dist[k] = 0
	q := []int{k}
	vis[k] = true
	for len(q) > 0 {
		u := q[0]
		q = q[1:]
		vis[u] = false
		for _, ne := range g[u] {
			v, w := ne[0], ne[1]
			if dist[v] > dist[u]+w {
				dist[v] = dist[u] + w
				if !vis[v] {
					q = append(q, v)
					vis[v] = true
				}
			}
		}
	}
	ans := 0
	for _, v := range dist {
		ans = max(ans, v)
	}
	if ans == inf {
		return -1
	}
	return ans
}

func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

C++

朴素 Dijkstra 算法：

class Solution {
public:
    const int inf = 0x3f3f;

    int networkDelayTime(vector<vector<int>>& times, int n, int k) {
        vector<vector<int>> g(n, vector<int>(n, inf));
        for (auto& t : times) g[t[0] - 1][t[1] - 1] = t[2];
        vector<bool> vis(n);
        vector<int> dist(n, inf);
        dist[k - 1] = 0;
        for (int i = 0; i < n; ++i)
        {
            int t = -1;
            for (int j = 0; j < n; ++j)
            {
                if (!vis[j] && (t == -1 || dist[t] > dist[j]))
                {
                    t = j;
                }
            }
            vis[t] = true;
            for (int j = 0; j < n; ++j)
            {
                dist[j] = min(dist[j], dist[t] + g[t][j]);
            }
        }
        int ans = *max_element(dist.begin(), dist.end());
        return ans == inf ? -1 : ans;
    }
};

堆优化 Dijkstra 算法：

class Solution {
public:
    const int inf = 0x3f3f;

    int networkDelayTime(vector<vector<int>>& times, int n, int k) {
        vector<vector<vector<int>>> g(n);
        for (auto& t : times) g[t[0] - 1].push_back({t[1] - 1, t[2]});
        vector<int> dist(n, inf);
        dist[k - 1] = 0;
        priority_queue<vector<int>, vector<vector<int>>, greater<vector<int>>> q;
        q.push({0, k - 1});
        while (!q.empty())
        {
            auto p = q.top();
            q.pop();
            int u = p[1];
            for (auto& ne : g[u])
            {
                int v = ne[0], w = ne[1];
                if (dist[v] > dist[u] + w)
                {
                    dist[v] = dist[u] + w;
                    q.push({dist[v], v});
                }
            }
        }
        int ans = *max_element(dist.begin(), dist.end());
        return ans == inf ? -1 : ans;
    }
};

Bellman Ford 算法：

class Solution {
public:
    int inf = 0x3f3f;

    int networkDelayTime(vector<vector<int>>& times, int n, int k) {
        vector<int> dist(n, inf);
        dist[k - 1] = 0;
        for (int i = 0; i < n; ++i)
        {
            vector<int> backup = dist;
            for (auto& e : times)
            {
                int u = e[0] - 1, v = e[1] - 1, w = e[2];
                dist[v] = min(dist[v], backup[u] + w);
            }
        }
        int ans = *max_element(dist.begin(), dist.end());
        return ans == inf ? -1 : ans;
    }
};

SPFA 算法：

class Solution {
public:
    const int inf = 0x3f3f;

    int networkDelayTime(vector<vector<int>>& times, int n, int k) {
        vector<int> dist(n, inf);
        vector<vector<vector<int>>> g(n);
        for (auto& e : times)
        {
            int u = e[0] - 1, v = e[1] - 1, w = e[2];
            g[u].push_back({v, w});
        }
        vector<bool> vis(n);
        --k;
        queue<int> q{{k}};
        vis[k] = true;
        dist[k] = 0;
        while (!q.empty())
        {
            int u = q.front();
            q.pop();
            vis[u] = false;
            for (auto& ne : g[u])
            {
                int v = ne[0], w = ne[1];
                if (dist[v] > dist[u] + w)
                {
                    dist[v] = dist[u] + w;
                    if (!vis[v])
                    {
                        q.push(v);
                        vis[v] = true;
                    }
                }
            }
        }
        int ans = *max_element(dist.begin(), dist.end());
        return ans == inf ? -1 : ans;
    }
};

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