diff --git a/solution/1300-1399/1334.Find the City With the Smallest Number of Neighbors at a Threshold Distance/README.md b/solution/1300-1399/1334.Find the City With the Smallest Number of Neighbors at a Threshold Distance/README.md index 2746c00439dc1..e27cac6344bc2 100644 --- a/solution/1300-1399/1334.Find the City With the Smallest Number of Neighbors at a Threshold Distance/README.md +++ b/solution/1300-1399/1334.Find the City With the Smallest Number of Neighbors at a Threshold Distance/README.md @@ -70,6 +70,16 @@ 时间复杂度 $O(n^3)$，空间复杂度 $O(n^2)$。其中 $n$ 为城市个数。 +**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。 + ### **Python3** @@ -81,7 +91,7 @@ class Solution: def findTheCity( self, n: int, edges: List[List[int]], distanceThreshold: int ) -> int: - def dijkstra(u): + def dijkstra(u: int) -> int: dist = [inf] * n dist[u] = 0 vis = [False] * n @@ -92,19 +102,43 @@ class Solution: k = j vis[k] = True for j in range(n): - dist[j] = min(dist[j], dist[k] + g[k][j]) + # 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 +``` - ans = n - t = inf +```python +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): - if (cnt := dijkstra(i)) < t: - t = cnt - ans = i + t = sum(d <= distanceThreshold for d in g[i]) + if t < cnt: + cnt, ans = t, i return ans ``` @@ -118,7 +152,7 @@ class Solution { private int[][] g; private int[] dist; private boolean[] vis; - private int inf = 1 << 30; + private final int inf = 1 << 30; private int distanceThreshold; public int findTheCity(int n, int[][] edges, int distanceThreshold) { @@ -135,11 +169,11 @@ class Solution { g[f][t] = w; g[t][f] = w; } - int ans = n, t = inf; + int ans = n, cnt = inf; for (int i = n - 1; i >= 0; --i) { - int cnt = dijkstra(i); - if (t > cnt) { - t = cnt; + int t = dijkstra(i); + if (t < cnt) { + cnt = t; ans = i; } } @@ -173,23 +207,62 @@ class Solution { } ``` +```java +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; + } +} +``` + ### **C++** ```cpp class Solution { public: int findTheCity(int n, vector>& edges, int distanceThreshold) { - const int inf = 1e7; - vector> g(n, vector(n, inf)); - vector dist(n, inf); - vector vis(n); + 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) { - dist.assign(n, inf); - vis.assign(n, false); + memset(dist, 0x3f, sizeof(dist)); + memset(vis, 0, sizeof(vis)); dist[u] = 0; for (int i = 0; i < n; ++i) { int k = -1; @@ -203,17 +276,44 @@ public: dist[j] = min(dist[j], dist[k] + g[k][j]); } } - int cnt = 0; - for (int& d : dist) { - cnt += d <= distanceThreshold; - } - return cnt; + return count_if(dist, dist + n, [&](int d) { return d <= distanceThreshold; }); }; - int ans = n, t = inf; + 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; + } +}; +``` + +```cpp +class Solution { +public: + int findTheCity(int n, vector>& 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 cnt = dijkstra(i); - if (t > cnt) { - t = cnt; + int t = count_if(g[i], g[i] + n, [&](int x) { return x <= distanceThreshold; }); + if (t < cnt) { + cnt = t; ans = i; } } @@ -241,7 +341,6 @@ func findTheCity(n int, edges [][]int, distanceThreshold int) int { g[f][t], g[t][f] = w, w } - ans, t := n, inf dijkstra := func(u int) (cnt int) { for i := range vis { vis[i] = false @@ -267,10 +366,11 @@ func findTheCity(n int, edges [][]int, distanceThreshold int) int { } return } + + ans, cnt := n, inf for i := n - 1; i >= 0; i-- { - cnt := dijkstra(i) - if t > cnt { - t = cnt + if t := dijkstra(i); t < cnt { + cnt = t ans = i } } @@ -278,6 +378,120 @@ func findTheCity(n int, edges [][]int, distanceThreshold int) int { } ``` +```go +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 +} +``` + +### **TypeScript** + +```ts +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; +} +``` + +```ts +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; +} +``` + ### **...** ``` diff --git a/solution/1300-1399/1334.Find the City With the Smallest Number of Neighbors at a Threshold Distance/README_EN.md b/solution/1300-1399/1334.Find the City With the Smallest Number of Neighbors at a Threshold Distance/README_EN.md index 6668ca19f6b40..726e7788daaad 100644 --- a/solution/1300-1399/1334.Find the City With the Smallest Number of Neighbors at a Threshold Distance/README_EN.md +++ b/solution/1300-1399/1334.Find the City With the Smallest Number of Neighbors at a Threshold Distance/README_EN.md @@ -63,7 +63,7 @@ class Solution: def findTheCity( self, n: int, edges: List[List[int]], distanceThreshold: int ) -> int: - def dijkstra(u): + def dijkstra(u: int) -> int: dist = [inf] * n dist[u] = 0 vis = [False] * n @@ -74,19 +74,43 @@ class Solution: k = j vis[k] = True for j in range(n): - dist[j] = min(dist[j], dist[k] + g[k][j]) + # 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 +``` + +```python +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 - ans = n - t = inf + 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): - if (cnt := dijkstra(i)) < t: - t = cnt - ans = i + t = sum(d <= distanceThreshold for d in g[i]) + if t < cnt: + cnt, ans = t, i return ans ``` @@ -98,7 +122,7 @@ class Solution { private int[][] g; private int[] dist; private boolean[] vis; - private int inf = 1 << 30; + private final int inf = 1 << 30; private int distanceThreshold; public int findTheCity(int n, int[][] edges, int distanceThreshold) { @@ -115,11 +139,11 @@ class Solution { g[f][t] = w; g[t][f] = w; } - int ans = n, t = inf; + int ans = n, cnt = inf; for (int i = n - 1; i >= 0; --i) { - int cnt = dijkstra(i); - if (t > cnt) { - t = cnt; + int t = dijkstra(i); + if (t < cnt) { + cnt = t; ans = i; } } @@ -153,23 +177,62 @@ class Solution { } ``` +```java +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; + } +} +``` + ### **C++** ```cpp class Solution { public: int findTheCity(int n, vector>& edges, int distanceThreshold) { - const int inf = 1e7; - vector> g(n, vector(n, inf)); - vector dist(n, inf); - vector vis(n); + 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) { - dist.assign(n, inf); - vis.assign(n, false); + memset(dist, 0x3f, sizeof(dist)); + memset(vis, 0, sizeof(vis)); dist[u] = 0; for (int i = 0; i < n; ++i) { int k = -1; @@ -183,17 +246,44 @@ public: dist[j] = min(dist[j], dist[k] + g[k][j]); } } - int cnt = 0; - for (int& d : dist) { - cnt += d <= distanceThreshold; - } - return cnt; + return count_if(dist, dist + n, [&](int d) { return d <= distanceThreshold; }); }; - int ans = n, t = inf; + int ans = n, cnt = n + 1; for (int i = n - 1; ~i; --i) { - int cnt = dijkstra(i); - if (t > cnt) { - t = cnt; + int t = dijkstra(i); + if (t < cnt) { + cnt = t; + ans = i; + } + } + return ans; + } +}; +``` + +```cpp +class Solution { +public: + int findTheCity(int n, vector>& 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; } } @@ -221,7 +311,6 @@ func findTheCity(n int, edges [][]int, distanceThreshold int) int { g[f][t], g[t][f] = w, w } - ans, t := n, inf dijkstra := func(u int) (cnt int) { for i := range vis { vis[i] = false @@ -247,10 +336,11 @@ func findTheCity(n int, edges [][]int, distanceThreshold int) int { } return } + + ans, cnt := n, inf for i := n - 1; i >= 0; i-- { - cnt := dijkstra(i) - if t > cnt { - t = cnt + if t := dijkstra(i); t < cnt { + cnt = t ans = i } } @@ -258,6 +348,120 @@ func findTheCity(n int, edges [][]int, distanceThreshold int) int { } ``` +```go +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 +} +``` + +### **TypeScript** + +```ts +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; +} +``` + +```ts +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; +} +``` + ### **...** ``` diff --git a/solution/1300-1399/1334.Find the City With the Smallest Number of Neighbors at a Threshold Distance/Solution.cpp b/solution/1300-1399/1334.Find the City With the Smallest Number of Neighbors at a Threshold Distance/Solution.cpp index d6b8a7ee52d64..e906580688985 100644 --- a/solution/1300-1399/1334.Find the City With the Smallest Number of Neighbors at a Threshold Distance/Solution.cpp +++ b/solution/1300-1399/1334.Find the City With the Smallest Number of Neighbors at a Threshold Distance/Solution.cpp @@ -1,44 +1,28 @@ -class Solution { -public: - int findTheCity(int n, vector>& edges, int distanceThreshold) { - const int inf = 1e7; - vector> g(n, vector(n, inf)); - vector dist(n, inf); - vector vis(n); - 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) { - dist.assign(n, inf); - vis.assign(n, 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[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]); - } - } - int cnt = 0; - for (int& d : dist) { - cnt += d <= distanceThreshold; - } - return cnt; - }; - int ans = n, t = inf; - for (int i = n - 1; ~i; --i) { - int cnt = dijkstra(i); - if (t > cnt) { - t = cnt; - ans = i; - } - } - return ans; - } +class Solution { +public: + int findTheCity(int n, vector>& 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; + } }; \ No newline at end of file diff --git a/solution/1300-1399/1334.Find the City With the Smallest Number of Neighbors at a Threshold Distance/Solution.go b/solution/1300-1399/1334.Find the City With the Smallest Number of Neighbors at a Threshold Distance/Solution.go index f306ba6aa63dd..6ed6f0b5fb6e8 100644 --- a/solution/1300-1399/1334.Find the City With the Smallest Number of Neighbors at a Threshold Distance/Solution.go +++ b/solution/1300-1399/1334.Find the City With the Smallest Number of Neighbors at a Threshold Distance/Solution.go @@ -1,7 +1,5 @@ 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) @@ -9,43 +7,33 @@ func findTheCity(n int, edges [][]int, distanceThreshold int) int { 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 } - ans, t := n, inf - dijkstra := func(u int) (cnt int) { - for i := range vis { - vis[i] = false - dist[i] = inf - } - dist[u] = 0 + for k := 0; k < n; k++ { + g[k][k] = 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++ + g[i][j] = min(g[i][j], g[i][k]+g[k][j]) } } - return } + + ans, cnt := n, n+1 for i := n - 1; i >= 0; i-- { - cnt := dijkstra(i) - if t > cnt { - t = cnt - ans = i + t := 0 + for _, x := range g[i] { + if x <= distanceThreshold { + t++ + } + } + if t < cnt { + cnt, ans = t, i } } + return ans } \ No newline at end of file diff --git a/solution/1300-1399/1334.Find the City With the Smallest Number of Neighbors at a Threshold Distance/Solution.java b/solution/1300-1399/1334.Find the City With the Smallest Number of Neighbors at a Threshold Distance/Solution.java index c0b8e940c7a1e..eb14f88e7388a 100644 --- a/solution/1300-1399/1334.Find the City With the Smallest Number of Neighbors at a Threshold Distance/Solution.java +++ b/solution/1300-1399/1334.Find the City With the Smallest Number of Neighbors at a Threshold Distance/Solution.java @@ -1,58 +1,36 @@ -class Solution { - private int n; - private int[][] g; - private int[] dist; - private boolean[] vis; - private 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, t = inf; - for (int i = n - 1; i >= 0; --i) { - int cnt = dijkstra(i); - if (t > cnt) { - t = cnt; - 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, 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; + } } \ No newline at end of file diff --git a/solution/1300-1399/1334.Find the City With the Smallest Number of Neighbors at a Threshold Distance/Solution.py b/solution/1300-1399/1334.Find the City With the Smallest Number of Neighbors at a Threshold Distance/Solution.py index 43fedc7630bd0..d0a78185a08f2 100644 --- a/solution/1300-1399/1334.Find the City With the Smallest Number of Neighbors at a Threshold Distance/Solution.py +++ b/solution/1300-1399/1334.Find the City With the Smallest Number of Neighbors at a Threshold Distance/Solution.py @@ -1,29 +1,22 @@ -class Solution: - def findTheCity( - self, n: int, edges: List[List[int]], distanceThreshold: int - ) -> int: - def dijkstra(u): - 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]) - 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 = n - t = inf - for i in range(n - 1, -1, -1): - if (cnt := dijkstra(i)) < t: - t = cnt - ans = i - return ans +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 diff --git a/solution/1300-1399/1334.Find the City With the Smallest Number of Neighbors at a Threshold Distance/Solution.ts b/solution/1300-1399/1334.Find the City With the Smallest Number of Neighbors at a Threshold Distance/Solution.ts new file mode 100644 index 0000000000000..20028a2e9c42e --- /dev/null +++ b/solution/1300-1399/1334.Find the City With the Smallest Number of Neighbors at a Threshold Distance/Solution.ts @@ -0,0 +1,25 @@ +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; +}