题目描述

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样例

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算法1

(暴力枚举) $O(n^2)$

blablabla

时间复杂度

参考文献

C++ 代码

#include<iostream>
#include<queue>
#include<cstring>

using namespace std;
const int N = 1010;
char a[N][N];
int dist[N][N];
int n, m;

void bfs()
{
    int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};
    memset(dist, -1, sizeof dist);
    queue<pair<int, int>> q;
    for (int i = 0; i < n; i ++)
        for (int j = 0; j < m; j ++)
            if (a[i][j] == '1')
            {
                q.push({i, j});
                dist[i][j] = 0;
            }

    while(q.size())
    {
        auto t = q.front();
        q.pop();
        int x = t.first, y = t.second;
        int d = dist[x][y];
        for (int i = 0; i < 4; i ++)
        {
            int a = x + dx[i], b = y +dy[i];
            if (a < 0 || a >= n || b < 0 || b >= m) continue;
            if(dist[a][b] != -1) continue;
            q.push({a, b});
            dist[a][b] = d + 1;
        }
    }
}

int main()
{
    cin >> n >> m;
    for (int i = 0; i < n; i ++)
        for (int j = 0; j < m; j ++)
            cin >> a[i][j];
    bfs();
    for (int i = 0; i < n; i ++)
    {
        for (int j = 0; j < m; j ++)
            cout << dist[i][j] << " ";
        cout << endl;
    }
    return 0;
}

算法2

(暴力枚举) $O(n^2)$

blablabla

时间复杂度

参考文献

C++ 代码

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