A virus is spreading rapidly, and your task is to quarantine the infected area by installing walls.

The world is modeled as a 2-D array of cells, where 0 represents uninfected cells, and 1 represents cells contaminated with the virus. A wall (and only one wall) can be installed between any two 4-directionally adjacent cells, on the shared boundary.

Every night, the virus spreads to all neighboring cells in all four directions unless blocked by a wall. Resources are limited. Each day, you can install walls around only one region -- the affected area (continuous block of infected cells) that threatens the most uninfected cells the following night. There will never be a tie.

Can you save the day? If so, what is the number of walls required? If not, and the world becomes fully infected, return the number of walls used.

Example 1:

Input: grid =

[[0,1,0,0,0,0,0,1],

 [0,1,0,0,0,0,0,1],

 [0,0,0,0,0,0,0,1],

 [0,0,0,0,0,0,0,0]]

Output: 10

Explanation:

There are 2 contaminated regions.

On the first day, add 5 walls to quarantine the viral region on the left. The board after the virus spreads is:

[[0,1,0,0,0,0,1,1],

 [0,1,0,0,0,0,1,1],

 [0,0,0,0,0,0,1,1],

 [0,0,0,0,0,0,0,1]]

On the second day, add 5 walls to quarantine the viral region on the right. The virus is fully contained.

Example 2:

Input: grid =

[[1,1,1],

 [1,0,1],

 [1,1,1]]

Output: 4

Explanation: Even though there is only one cell saved, there are 4 walls built.

Notice that walls are only built on the shared boundary of two different cells.

Example 3:

Input: grid =

[[1,1,1,0,0,0,0,0,0],

 [1,0,1,0,1,1,1,1,1],

 [1,1,1,0,0,0,0,0,0]]

Output: 13

Explanation: The region on the left only builds two new walls.

Note:

The number of rows and columns of grid will each be in the range [1, 50].
Each grid[i][j] will be either 0 or 1.
Throughout the described process, there is always a contiguous viral region that will infect strictly more uncontaminated squares in the next round.

Python:

class Solution(object):

    def containVirus(self, grid):

        """

        :type grid: List[List[int]]

        :rtype: int

        """

        directions = [(0, 1), (0, -1), (-1, 0), (1, 0)]

        def dfs(grid, r, c, lookup, regions, frontiers, perimeters):

            if (r, c) in lookup:

                return

            lookup.add((r, c))

            regions[-1].add((r, c))

            for d in directions:

                nr, nc = r+d[0], c+d[1]

                if not (0 <= nr < len(grid) and \

                        0 <= nc < len(grid[r])):

                    continue

                if grid[nr][nc] == 1:

                    dfs(grid, nr, nc, lookup, regions, frontiers, perimeters)

                elif grid[nr][nc] == 0:

                    frontiers[-1].add((nr, nc))

                    perimeters[-1] += 1

        result = 0

        while True:

            lookup, regions, frontiers, perimeters = set(), [], [], []

            for r, row in enumerate(grid):

                for c, val in enumerate(row):

                    if val == 1 and (r, c) not in lookup:

                        regions.append(set())

                        frontiers.append(set())

                        perimeters.append(0)

                        dfs(grid, r, c, lookup, regions, frontiers, perimeters)

            if not regions: break

            triage_idx = frontiers.index(max(frontiers, key = len))

            for i, region in enumerate(regions):

                if i == triage_idx:

                    result += perimeters[i]

                    for r, c in region:

                        grid[r][c] = -1

                    continue

                for r, c in region:

                    for d in directions:

                        nr, nc = r+d[0], c+d[1]

                        if not (0 <= nr < len(grid) and \

                                0 <= nc < len(grid[r])):

                            continue

                        if grid[nr][nc] == 0:

                            grid[nr][nc] = 1

        return result

C++:

class Solution {

public:

    int containVirus(vector<vector<int>>& grid) {

        int res = 0, m = grid.size(), n = grid[0].size();

        vector<vector<int>> dirs{{-1,0},{0,1},{1,0},{0,-1}};

        while (true) {

            unordered_set<int> visited;

            vector<vector<vector<int>>> all;

            for (int i = 0; i < m; ++i) {

                for (int j = 0; j < n; ++j) {

                    if (grid[i][j] == 1 && !visited.count(i * n + j)) {

                        queue<int> q{{i * n + j}};

                        vector<int> virus{i * n + j};

                        vector<int> walls;

                        visited.insert(i * n + j);

                        while (!q.empty()) {

                            auto t = q.front(); q.pop();

                            for (auto dir : dirs) {

                                int x = (t / n) + dir[0], y = (t % n) + dir[1];

                                if (x < 0 || x >= m || y < 0 || y >= n || visited.count(x * n + y)) continue;

                                if (grid[x][y] == -1) continue;

                                else if (grid[x][y] == 0) walls.push_back(x * n + y);

                                else if (grid[x][y] == 1) {

                                    visited.insert(x * n + y);

                                    virus.push_back(x * n + y);

                                    q.push(x * n + y);

                                }

                            }

                        }

                        unordered_set<int> s(walls.begin(), walls.end());

                        vector<int> cells{(int)s.size()};

                        all.push_back({cells ,walls, virus});

                    }

                }

            }

            if (all.empty()) break;

            sort(all.begin(), all.end(), [](vector<vector<int>> &a, vector<vector<int>> &b) {return a[0][0] > b[0][0];});

            for (int i = 0; i < all.size(); ++i) {

                if (i == 0) {

                    vector<int> virus = all[0][2];

                    for (int idx : virus) grid[idx / n][idx % n] = -1;

                    res += all[0][1].size();

                } else {

                    vector<int> wall = all[i][1];

                    for (int idx : wall) grid[idx / n][idx % n] = 1;

                }

            }

        }

        return res;

    }

};

All LeetCode Questions List 题目汇总

巴特西

[LeetCode] 749. Contain Virus 包含病毒

All LeetCode Questions List 题目汇总

最新文章

热门文章