2492. Minimum Score of a Path Between Two Cities

Medium
You are given a positive integer n representing n cities numbered from 1 to n. You are also given a 2D array roads where roads[i] = [ai, bi, distancei] indicates that there is a bidirectional road between cities ai and bi with a distance equal to distancei. The cities graph is not necessarily connected.
The score of a path between two cities is defined as the minimum distance of a road in this path.
Return the minimum possible score of a path between cities 1 and n.
Note:
  • A path is a sequence of roads between two cities.
  • It is allowed for a path to contain the same road multiple times, and you can visit cities 1 and n multiple times along the path.
  • The test cases are generated such that there is at least one path between 1 and n.
Example 1:
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Input: n = 4, roads = [[1,2,9],[2,3,6],[2,4,5],[1,4,7]]
Output: 5
Explanation: The path from city 1 to 4 with the minimum score is: 1 -> 2 -> 4. The score of this path is min(9,5) = 5.
It can be shown that no other path has less score.
Example 2:
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Input: n = 4, roads = [[1,2,2],[1,3,4],[3,4,7]]
Output: 2
Explanation: The path from city 1 to 4 with the minimum score is: 1 -> 2 -> 1 -> 3 -> 4. The score of this path is min(2,2,4,7) = 2.
Constraints:
  • 2 <= n <= 10^5
  • 1 <= roads.length <= 10^5
  • roads[i].length == 3
  • 1 <= ai, bi <= n
  • ai != bi
  • 1 <= distancei <= 10^4
  • There are no repeated edges.
  • There is at least one path between 1 and n.

解題

使用 queue 來做 BFS
func minScore(n int, roads [][]int) int {
adj := make([][][]int, n+1)
for i:=0; i<=n; i++ {
adj[i] = [][]int{}
}
​
for _, road := range roads {
adj[road[0]] = append(adj[road[0]], []int{ road[1], road[2] })
adj[road[1]] = append(adj[road[1]], []int{ road[0], road[2] })
}
​
ans := 100000000
queue := make([][]int, 0)
queue = append(queue, []int{1, 100000000})
​
visited := make([]bool, n+1)
visited[1] = true
​
for len(queue) != 0 {
ct := queue[0][0]
queue = queue[1:]
for _, a := range adj[ct] {
ans = min(ans, a[1])
if !visited[a[0]] {
visited[a[0]] = true
queue = append(queue, []int{ a[0], a[1]})
}
}
}
​
return ans
}
​
func min(a, b int) int {
if a < b { return a }
return b
}
​
Union-find
func minScore(n int, roads [][]int) int {
group := make([]int, n+1)
minpath := make([]int, n+1)
​
var find func(int) int
find = func(a int) int {
if a == group[a] { return a }
return find(group[a])
}
​
var union func(int, int, int)
union = func(a int, b int, dis int) {
aroot := find(a)
broot := find(b)
​
if broot < aroot {
group[broot] = aroot
minpath[aroot] = min(dis, min(minpath[aroot], minpath[broot]))
} else {
group[aroot] = broot
minpath[broot] = min(dis, min(minpath[broot], minpath[aroot]))
}
}
​
for i:=0; i<=n; i++ {
group[i] = i
minpath[i] = 10000000
}
​
for _, road := range roads {
union(road[0], road[1], road[2])
}
​
return minpath[find(n)]
}
​
func min(a, b int) int {
if a < b { return a }
return b
}
​
​