Question

1334. 阈值距离内邻居最少的城市

English Version

题目描述

有 n 个城市，按从 0 到 n-1 编号。给你一个边数组 edges，其中 edges[i] = [fromi, toi, weighti] 代表 fromi 和 toi 两个城市之间的双向加权边，距离阈值是一个整数 distanceThreshold。

返回能通过某些路径到达其他城市数目最少、且路径距离 最大 为 distanceThreshold 的城市。如果有多个这样的城市，则返回编号最大的城市。

注意，连接城市 _i_ 和 _j_ 的路径的距离等于沿该路径的所有边的权重之和。

 

示例 1：

输入：n = 4, edges = [[0,1,3],[1,2,1],[1,3,4],[2,3,1]], distanceThreshold = 4
输出：3
解释：城市分布图如上。
每个城市阈值距离 distanceThreshold = 4 内的邻居城市分别是：
城市 0 -> [城市 1, 城市 2] 
城市 1 -> [城市 0, 城市 2, 城市 3] 
城市 2 -> [城市 0, 城市 1, 城市 3] 
城市 3 -> [城市 1, 城市 2] 
城市 0 和 3 在阈值距离 4 以内都有 2 个邻居城市，但是我们必须返回城市 3，因为它的编号最大。

示例 2：

输入：n = 5, edges = [[0,1,2],[0,4,8],[1,2,3],[1,4,2],[2,3,1],[3,4,1]], distanceThreshold = 2
输出：0
解释：城市分布图如上。 
每个城市阈值距离 distanceThreshold = 2 内的邻居城市分别是：
城市 0 -> [城市 1] 
城市 1 -> [城市 0, 城市 4] 
城市 2 -> [城市 3, 城市 4] 
城市 3 -> [城市 2, 城市 4]
城市 4 -> [城市 1, 城市 2, 城市 3] 
城市 0 在阈值距离 2 以内只有 1 个邻居城市。

 

提示：

• 2 <= n <= 100
• 1 <= edges.length <= n * (n - 1) / 2
• edges[i].length == 3
• 0 <= fromi < toi < n
• 1 <= weighti, distanceThreshold <= 10^4
• 所有 (fromi, toi) 都是不同的。

解法

方法一：Dijkstra 算法

我们可以枚举每个城市 $i$ 作为起点，使用 Dijkstra 算法求出从 $i$ 到其他城市的最短距离，然后统计距离不超过阈值的城市个数，最后取最小的个数且编号最大的城市。

时间复杂度 $O(n^3)$，空间复杂度 $O(n^2)$。其中 $n$ 为城市个数。

class Solution:
  def findTheCity(
    self, n: int, edges: List[List[int]], distanceThreshold: int
  ) -> int:
    def dijkstra(u: int) -> int:
      dist = [inf] * n
      dist[u] = 0
      vis = [False] * n
      for _ in range(n):
        k = -1
        for j in range(n):
          if not vis[j] and (k == -1 or dist[k] > dist[j]):
            k = j
        vis[k] = True
        for j in range(n):
          # dist[j] = min(dist[j], dist[k] + g[k][j])
          if dist[k] + g[k][j] < dist[j]:
            dist[j] = dist[k] + g[k][j]
      return sum(d <= distanceThreshold for d in dist)

    g = [[inf] * n for _ in range(n)]
    for f, t, w in edges:
      g[f][t] = g[t][f] = w
    ans, cnt = n, inf
    for i in range(n - 1, -1, -1):
      if (t := dijkstra(i)) < cnt:
        cnt, ans = t, i
    return ans

class Solution {
  private int n;
  private int[][] g;
  private int[] dist;
  private boolean[] vis;
  private final int inf = 1 << 30;
  private int distanceThreshold;

  public int findTheCity(int n, int[][] edges, int distanceThreshold) {
    this.n = n;
    this.distanceThreshold = distanceThreshold;
    g = new int[n][n];
    dist = new int[n];
    vis = new boolean[n];
    for (var e : g) {
      Arrays.fill(e, inf);
    }
    for (var e : edges) {
      int f = e[0], t = e[1], w = e[2];
      g[f][t] = w;
      g[t][f] = w;
    }
    int ans = n, cnt = inf;
    for (int i = n - 1; i >= 0; --i) {
      int t = dijkstra(i);
      if (t < cnt) {
        cnt = t;
        ans = i;
      }
    }
    return ans;
  }

  private int dijkstra(int u) {
    Arrays.fill(dist, inf);
    Arrays.fill(vis, false);
    dist[u] = 0;
    for (int i = 0; i < n; ++i) {
      int k = -1;
      for (int j = 0; j < n; ++j) {
        if (!vis[j] && (k == -1 || dist[k] > dist[j])) {
          k = j;
        }
      }
      vis[k] = true;
      for (int j = 0; j < n; ++j) {
        dist[j] = Math.min(dist[j], dist[k] + g[k][j]);
      }
    }
    int cnt = 0;
    for (int d : dist) {
      if (d <= distanceThreshold) {
        ++cnt;
      }
    }
    return cnt;
  }
}

class Solution {
public:
  int findTheCity(int n, vector<vector<int>>& edges, int distanceThreshold) {
    int g[n][n];
    int dist[n];
    bool vis[n];
    memset(g, 0x3f, sizeof(g));
    for (auto& e : edges) {
      int f = e[0], t = e[1], w = e[2];
      g[f][t] = g[t][f] = w;
    }
    auto dijkstra = [&](int u) {
      memset(dist, 0x3f, sizeof(dist));
      memset(vis, 0, sizeof(vis));
      dist[u] = 0;
      for (int i = 0; i < n; ++i) {
        int k = -1;
        for (int j = 0; j < n; ++j) {
          if (!vis[j] && (k == -1 || dist[j] < dist[k])) {
            k = j;
          }
        }
        vis[k] = true;
        for (int j = 0; j < n; ++j) {
          dist[j] = min(dist[j], dist[k] + g[k][j]);
        }
      }
      return count_if(dist, dist + n, [&](int d) { return d <= distanceThreshold; });
    };
    int ans = n, cnt = n + 1;
    for (int i = n - 1; ~i; --i) {
      int t = dijkstra(i);
      if (t < cnt) {
        cnt = t;
        ans = i;
      }
    }
    return ans;
  }
};

func findTheCity(n int, edges [][]int, distanceThreshold int) int {
  g := make([][]int, n)
  dist := make([]int, n)
  vis := make([]bool, n)
  const inf int = 1e7
  for i := range g {
    g[i] = make([]int, n)
    for j := range g[i] {
      g[i][j] = inf
    }
  }
  for _, e := range edges {
    f, t, w := e[0], e[1], e[2]
    g[f][t], g[t][f] = w, w
  }

  dijkstra := func(u int) (cnt int) {
    for i := range vis {
      vis[i] = false
      dist[i] = inf
    }
    dist[u] = 0
    for i := 0; i < n; i++ {
      k := -1
      for j := 0; j < n; j++ {
        if !vis[j] && (k == -1 || dist[j] < dist[k]) {
          k = j
        }
      }
      vis[k] = true
      for j := 0; j < n; j++ {
        dist[j] = min(dist[j], dist[k]+g[k][j])
      }
    }
    for _, d := range dist {
      if d <= distanceThreshold {
        cnt++
      }
    }
    return
  }

  ans, cnt := n, inf
  for i := n - 1; i >= 0; i-- {
    if t := dijkstra(i); t < cnt {
      cnt = t
      ans = i
    }
  }
  return ans
}

function findTheCity(n: number, edges: number[][], distanceThreshold: number): number {
  const g: number[][] = Array.from({ length: n }, () => Array(n).fill(Infinity));
  const dist: number[] = Array(n).fill(Infinity);
  const vis: boolean[] = Array(n).fill(false);
  for (const [f, t, w] of edges) {
    g[f][t] = g[t][f] = w;
  }

  const dijkstra = (u: number): number => {
    dist.fill(Infinity);
    vis.fill(false);
    dist[u] = 0;
    for (let i = 0; i < n; ++i) {
      let k = -1;
      for (let j = 0; j < n; ++j) {
        if (!vis[j] && (k === -1 || dist[j] < dist[k])) {
          k = j;
        }
      }
      vis[k] = true;
      for (let j = 0; j < n; ++j) {
        dist[j] = Math.min(dist[j], dist[k] + g[k][j]);
      }
    }
    return dist.filter(d => d <= distanceThreshold).length;
  };

  let ans = n;
  let cnt = Infinity;
  for (let i = n - 1; i >= 0; --i) {
    const t = dijkstra(i);
    if (t < cnt) {
      cnt = t;
      ans = i;
    }
  }

  return ans;
}

方法二：Floyd 算法

我们定义 $g[i][j]$ 表示城市 $i$ 到城市 $j$ 的最短距离，初始时 $g[i][j] = \infty$, $g[i][i] = 0$，然后我们遍历所有边，对于每条边 $(f, t, w)$，我们令 $g[f][t] = g[t][f] = w$。

接下来，我们用 Floyd 算法求出任意两点之间的最短距离。具体地，我们先枚举中间点 $k$，再枚举起点 $i$ 和终点 $j$，如果 $g[i][k] + g[k][j] \lt g[i][j]$，那么我们就用更短的距离 $g[i][k] + g[k][j]$ 更新 $g[i][j]$。

最后，我们枚举每个城市 $i$ 作为起点，统计距离不超过阈值的城市个数，最后取最小的个数且编号最大的城市。

时间复杂度 $O(n^3)$，空间复杂度 $O(n^2)$。其中 $n$ 为城市个数。

class Solution:
  def findTheCity(
    self, n: int, edges: List[List[int]], distanceThreshold: int
  ) -> int:
    g = [[inf] * n for _ in range(n)]
    for f, t, w in edges:
      g[f][t] = g[t][f] = w

    for k in range(n):
      g[k][k] = 0
      for i in range(n):
        for j in range(n):
          # g[i][j] = min(g[i][j], g[i][k] + g[k][j])
          if g[i][k] + g[k][j] < g[i][j]:
            g[i][j] = g[i][k] + g[k][j]

    ans, cnt = n, inf
    for i in range(n - 1, -1, -1):
      t = sum(d <= distanceThreshold for d in g[i])
      if t < cnt:
        cnt, ans = t, i
    return ans

class Solution {
  public int findTheCity(int n, int[][] edges, int distanceThreshold) {
    final int inf = 1 << 29;
    int[][] g = new int[n][n];
    for (var e : g) {
      Arrays.fill(e, inf);
    }
    for (var e : edges) {
      int f = e[0], t = e[1], w = e[2];
      g[f][t] = w;
      g[t][f] = w;
    }
    for (int k = 0; k < n; ++k) {
      g[k][k] = 0;
      for (int i = 0; i < n; ++i) {
        for (int j = 0; j < n; ++j) {
          g[i][j] = Math.min(g[i][j], g[i][k] + g[k][j]);
        }
      }
    }
    int ans = n, cnt = inf;
    for (int i = n - 1; i >= 0; --i) {
      int t = 0;
      for (int d : g[i]) {
        if (d <= distanceThreshold) {
          ++t;
        }
      }
      if (t < cnt) {
        cnt = t;
        ans = i;
      }
    }
    return ans;
  }
}

class Solution {
public:
  int findTheCity(int n, vector<vector<int>>& edges, int distanceThreshold) {
    int g[n][n];
    memset(g, 0x3f, sizeof(g));
    for (auto& e : edges) {
      int f = e[0], t = e[1], w = e[2];
      g[f][t] = g[t][f] = w;
    }
    for (int k = 0; k < n; ++k) {
      g[k][k] = 0;
      for (int i = 0; i < n; ++i) {
        for (int j = 0; j < n; ++j) {
          g[i][j] = min(g[i][j], g[i][k] + g[k][j]);
        }
      }
    }
    int ans = n, cnt = n + 1;
    for (int i = n - 1; ~i; --i) {
      int t = count_if(g[i], g[i] + n, [&](int x) { return x <= distanceThreshold; });
      if (t < cnt) {
        cnt = t;
        ans = i;
      }
    }
    return ans;
  }
};

func findTheCity(n int, edges [][]int, distanceThreshold int) int {
  g := make([][]int, n)
  const inf int = 1e7
  for i := range g {
    g[i] = make([]int, n)
    for j := range g[i] {
      g[i][j] = inf
    }
  }

  for _, e := range edges {
    f, t, w := e[0], e[1], e[2]
    g[f][t], g[t][f] = w, w
  }

  for k := 0; k < n; k++ {
    g[k][k] = 0
    for i := 0; i < n; i++ {
      for j := 0; j < n; j++ {
        g[i][j] = min(g[i][j], g[i][k]+g[k][j])
      }
    }
  }

  ans, cnt := n, n+1
  for i := n - 1; i >= 0; i-- {
    t := 0
    for _, x := range g[i] {
      if x <= distanceThreshold {
        t++
      }
    }
    if t < cnt {
      cnt, ans = t, i
    }
  }

  return ans
}

function findTheCity(n: number, edges: number[][], distanceThreshold: number): number {
  const g: number[][] = Array.from({ length: n }, () => Array(n).fill(Infinity));
  for (const [f, t, w] of edges) {
    g[f][t] = g[t][f] = w;
  }
  for (let k = 0; k < n; ++k) {
    g[k][k] = 0;
    for (let i = 0; i < n; ++i) {
      for (let j = 0; j < n; ++j) {
        g[i][j] = Math.min(g[i][j], g[i][k] + g[k][j]);
      }
    }
  }

  let ans = n,
    cnt = n + 1;
  for (let i = n - 1; i >= 0; --i) {
    const t = g[i].filter(x => x <= distanceThreshold).length;
    if (t < cnt) {
      cnt = t;
      ans = i;
    }
  }
  return ans;
}