Question

924. 尽量减少恶意软件的传播

English Version

题目描述

给出了一个由 n 个节点组成的网络，用 n × n 个邻接矩阵图

 graph 表示。在节点网络中，当 graph[i][j] = 1 时，表示节点 i 能够直接连接到另一个节点 j。 

一些节点 initial 最初被恶意软件感染。只要两个节点直接连接，且其中至少一个节点受到恶意软件的感染，那么两个节点都将被恶意软件感染。这种恶意软件的传播将继续，直到没有更多的节点可以被这种方式感染。

假设 M(initial) 是在恶意软件停止传播之后，整个网络中感染恶意软件的最终节点数。

如果从 initial 中移除某一节点能够最小化 M(initial)， 返回该节点。如果有多个节点满足条件，就返回索引最小的节点。

请注意，如果某个节点已从受感染节点的列表 initial 中删除，它以后仍有可能因恶意软件传播而受到感染。

 

示例 1：

输入：graph = [[1,1,0],[1,1,0],[0,0,1]], initial = [0,1]
输出：0

示例 2：

输入：graph = [[1,0,0],[0,1,0],[0,0,1]], initial = [0,2]
输出：0

示例 3：

输入：graph = [[1,1,1],[1,1,1],[1,1,1]], initial = [1,2]
输出：1

 

提示：

• n == graph.length
• n == graph[i].length
• 2 <= n <= 300
• graph[i][j] == 0 或 1.
• graph[i][j] == graph[j][i]
• graph[i][i] == 1
• 1 <= initial.length <= n
• 0 <= initial[i] <= n - 1
• initial 中所有整数均不重复

解法

方法一

class Solution:
  def minMalwareSpread(self, graph: List[List[int]], initial: List[int]) -> int:
    n = len(graph)
    p = list(range(n))
    size = [1] * n

    def find(x):
      if p[x] != x:
        p[x] = find(p[x])
      return p[x]

    for i in range(n):
      for j in range(i + 1, n):
        if graph[i][j] == 1:
          pa, pb = find(i), find(j)
          if pa == pb:
            continue
          p[pa] = pb
          size[pb] += size[pa]

    mi = inf
    res = initial[0]
    initial.sort()
    for i in range(len(initial)):
      t = 0
      s = set()
      for j in range(len(initial)):
        if i == j:
          continue
        if find(initial[j]) in s:
          continue
        s.add(find(initial[j]))
        t += size[find(initial[j])]
      if mi > t:
        mi = t
        res = initial[i]
    return res

class Solution {
  private int[] p;

  public int minMalwareSpread(int[][] graph, int[] initial) {
    int n = graph.length;
    p = new int[n];
    for (int i = 0; i < n; ++i) {
      p[i] = i;
    }
    int[] size = new int[n];
    Arrays.fill(size, 1);
    for (int i = 0; i < n; ++i) {
      for (int j = i + 1; j < n; ++j) {
        if (graph[i][j] == 1) {
          int pa = find(i), pb = find(j);
          if (pa == pb) {
            continue;
          }
          p[pa] = pb;
          size[pb] += size[pa];
        }
      }
    }
    int mi = Integer.MAX_VALUE;
    int res = initial[0];
    Arrays.sort(initial);
    for (int i = 0; i < initial.length; ++i) {
      int t = 0;
      Set<Integer> s = new HashSet<>();
      for (int j = 0; j < initial.length; ++j) {
        if (i == j) {
          continue;
        }
        if (s.contains(find(initial[j]))) {
          continue;
        }
        s.add(find(initial[j]));
        t += size[find(initial[j])];
      }
      if (mi > t) {
        mi = t;
        res = initial[i];
      }
    }
    return res;
  }

  private int find(int x) {
    if (p[x] != x) {
      p[x] = find(p[x]);
    }
    return p[x];
  }
}

class Solution {
public:
  vector<int> p;

  int minMalwareSpread(vector<vector<int>>& graph, vector<int>& initial) {
    int n = graph.size();
    p.resize(n);
    for (int i = 0; i < n; ++i) p[i] = i;
    vector<int> size(n, 1);
    for (int i = 0; i < n; ++i) {
      for (int j = i + 1; j < n; ++j) {
        if (graph[i][j]) {
          int pa = find(i), pb = find(j);
          if (pa == pb) continue;
          p[pa] = pb;
          size[pb] += size[pa];
        }
      }
    }
    int mi = 400;
    int res = initial[0];
    sort(initial.begin(), initial.end());
    for (int i = 0; i < initial.size(); ++i) {
      int t = 0;
      unordered_set<int> s;
      for (int j = 0; j < initial.size(); ++j) {
        if (i == j) continue;
        if (s.count(find(initial[j]))) continue;
        s.insert(find(initial[j]));
        t += size[find(initial[j])];
      }
      if (mi > t) {
        mi = t;
        res = initial[i];
      }
    }
    return res;
  }

  int find(int x) {
    if (p[x] != x) p[x] = find(p[x]);
    return p[x];
  }
};

var p []int

func minMalwareSpread(graph [][]int, initial []int) int {
  n := len(graph)
  p = make([]int, n)
  size := make([]int, n)
  for i := 0; i < n; i++ {
    p[i] = i
    size[i] = 1
  }
  for i := 0; i < n; i++ {
    for j := i + 1; j < n; j++ {
      if graph[i][j] == 1 {
        pa, pb := find(i), find(j)
        if pa == pb {
          continue
        }
        p[pa] = pb
        size[pb] += size[pa]
      }
    }
  }
  mi := 400
  res := initial[0]
  sort.Ints(initial)
  for i := 0; i < len(initial); i++ {
    t := 0
    s := make(map[int]bool)
    for j := 0; j < len(initial); j++ {
      if i == j {
        continue
      }
      if s[find(initial[j])] {
        continue
      }
      s[find(initial[j])] = true
      t += size[find(initial[j])]
    }
    if mi > t {
      mi = t
      res = initial[i]
    }
  }
  return res
}

func find(x int) int {
  if p[x] != x {
    p[x] = find(p[x])
  }
  return p[x]
}