题目描述
给你一个由 n 个节点（下标从 0 开始）组成的无向加权图，该图由一个描述边的列表组成，其中 edges[i] = [a, b] 表示连接节点 a 和 b 的一条无向边，且该边遍历成功的概率为 succProb[i] 。
指定两个节点分别作为起点 start 和终点 end ，请你找出从起点到终点成功概率最大的路径，并返回其成功概率。
如果不存在从 start 到 end 的路径，请 返回 0 。只要答案与标准答案的误差不超过 1e-5 ，就会被视作正确答案。
示例 1：
输入：n = 3, edges = [[0,1],[1,2],[0,2]], succProb = [0.5,0.5,0.2], start = 0, end = 2
输出：0.25000
解释：从起点到终点有两条路径，其中一条的成功概率为 0.2 ，而另一条为 0.5 * 0.5 = 0.25
示例 2：
输入：n = 3, edges = [[0,1],[1,2],[0,2]], succProb = [0.5,0.5,0.3], start = 0, end = 2
输出：0.30000
示例 3：
输入：n = 3, edges = [[0,1]], succProb = [0.5], start = 0, end = 2
输出：0.00000
解释：节点 0 和 节点 2 之间不存在路径
提示：
2 <= n <= 10^40 <= start, end < nstart != end0 <= a, b < na != b0 <= succProb.length == edges.length <= 2*10^40 <= succProb[i] <= 1- 每两个节点之间最多有一条边
解法
方法一：堆优化 Dijkstra 算法
时间复杂度 O(mlogn)。
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26 | class Solution:
def maxProbability(
self,
n: int,
edges: List[List[int]],
succProb: List[float],
start: int,
end: int,
) -> float:
g = defaultdict(list)
for (a, b), s in zip(edges, succProb):
g[a].append((b, s))
g[b].append((a, s))
q = [(-1, start)]
d = [0] * n
d[start] = 1
while q:
w, u = heappop(q)
w = -w
if d[u] > w:
continue
for v, t in g[u]:
if d[v] < d[u] * t:
d[v] = d[u] * t
heappush(q, (-d[v], v))
return d[end]
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32 | class Solution {
public double maxProbability(int n, int[][] edges, double[] succProb, int start, int end) {
List<Pair<Integer, Double>>[] g = new List[n];
Arrays.setAll(g, k -> new ArrayList<>());
for (int i = 0; i < edges.length; ++i) {
int a = edges[i][0], b = edges[i][1];
double s = succProb[i];
g[a].add(new Pair<>(b, s));
g[b].add(new Pair<>(a, s));
}
PriorityQueue<Pair<Double, Integer>> q
= new PriorityQueue<>(Comparator.comparingDouble(Pair::getKey));
double[] d = new double[n];
d[start] = 1.0;
q.offer(new Pair<>(-1.0, start));
while (!q.isEmpty()) {
Pair<Double, Integer> p = q.poll();
double w = p.getKey();
w *= -1;
int u = p.getValue();
for (Pair<Integer, Double> ne : g[u]) {
int v = ne.getKey();
double t = ne.getValue();
if (d[v] < d[u] * t) {
d[v] = d[u] * t;
q.offer(new Pair<>(-d[v], v));
}
}
}
return d[end];
}
}
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32 | class Solution {
public:
double maxProbability(int n, vector<vector<int>>& edges, vector<double>& succProb, int start, int end) {
vector<vector<pair<int, double>>> g(n);
for (int i = 0; i < edges.size(); ++i) {
int a = edges[i][0], b = edges[i][1];
double s = succProb[i];
g[a].push_back({b, s});
g[b].push_back({a, s});
}
vector<double> d(n);
d[start] = 1.0;
queue<pair<double, int>> q;
q.push({1.0, start});
while (!q.empty()) {
auto p = q.front();
q.pop();
double w = p.first;
int u = p.second;
if (d[u] > w) continue;
for (auto& e : g[u]) {
int v = e.first;
double t = e.second;
if (d[v] < d[u] * t) {
d[v] = d[u] * t;
q.push({d[v], v});
}
}
}
return d[end];
}
};
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34 | func maxProbability(n int, edges [][]int, succProb []float64, start int, end int) float64 {
g := make([][]pair, n)
for i, e := range edges {
a, b, s := e[0], e[1], succProb[i]
g[a] = append(g[a], pair{b, s})
g[b] = append(g[b], pair{a, s})
}
d := make([]float64, n)
d[start] = 1
vis := make([]bool, n)
q := []int{start}
vis[start] = true
for len(q) > 0 {
i := q[0]
q = q[1:]
vis[i] = false
for _, ne := range g[i] {
j, s := ne.idx, ne.s
if d[j] < d[i]*s {
d[j] = d[i] * s
if !vis[j] {
q = append(q, j)
vis[j] = true
}
}
}
}
return d[end]
}
type pair struct {
idx int
s float64
}
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方法二：SPFA 算法
时间复杂度，平均情况下 O(m)，最坏情况下 O(nm)，n 表示点数，m 表示边数。
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28 | class Solution:
def maxProbability(
self,
n: int,
edges: List[List[int]],
succProb: List[float],
start: int,
end: int,
) -> float:
g = defaultdict(list)
for (a, b), s in zip(edges, succProb):
g[a].append((b, s))
g[b].append((a, s))
d = [0] * n
vis = [False] * n
d[start] = 1
q = deque([start])
vis[start] = True
while q:
i = q.popleft()
vis[i] = False
for j, s in g[i]:
if d[j] < d[i] * s:
d[j] = d[i] * s
if not vis[j]:
q.append(j)
vis[j] = True
return d[end]
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34 | class Solution {
public double maxProbability(int n, int[][] edges, double[] succProb, int start, int end) {
List<Pair<Integer, Double>>[] g = new List[n];
Arrays.setAll(g, k -> new ArrayList<>());
for (int i = 0; i < edges.length; ++i) {
int a = edges[i][0], b = edges[i][1];
double s = succProb[i];
g[a].add(new Pair<>(b, s));
g[b].add(new Pair<>(a, s));
}
double[] d = new double[n];
d[start] = 1.0;
boolean[] vis = new boolean[n];
Deque<Integer> q = new ArrayDeque<>();
q.offer(start);
vis[start] = true;
while (!q.isEmpty()) {
int i = q.poll();
vis[i] = false;
for (Pair<Integer, Double> ne : g[i]) {
int j = ne.getKey();
double s = ne.getValue();
if (d[j] < d[i] * s) {
d[j] = d[i] * s;
if (!vis[j]) {
q.offer(j);
vis[j] = true;
}
}
}
}
return d[end];
}
}
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34 | class Solution {
public:
double maxProbability(int n, vector<vector<int>>& edges, vector<double>& succProb, int start, int end) {
vector<vector<pair<int, double>>> g(n);
for (int i = 0; i < edges.size(); ++i) {
int a = edges[i][0], b = edges[i][1];
double s = succProb[i];
g[a].push_back({b, s});
g[b].push_back({a, s});
}
vector<double> d(n);
vector<bool> vis(n);
d[start] = 1.0;
queue<int> q{{start}};
vis[start] = true;
while (!q.empty()) {
int i = q.front();
q.pop();
vis[i] = false;
for (auto& ne : g[i]) {
int j = ne.first;
double s = ne.second;
if (d[j] < d[i] * s) {
d[j] = d[i] * s;
if (!vis[j]) {
q.push(j);
vis[j] = true;
}
}
}
}
return d[end];
}
};
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