hihocoder offer收割编程练习赛9 B 水陆距离

思路：

宽搜，多个起点。

实现：

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <cstring>
using namespace std;

const int INF = 0x3f3f3f3f;

char a[805][805];
int n, m, d[805][805];
bool vis[805][805];
int dx[4] = { 1, 0, -1, 0 };
int dy[4] = { 0, 1, 0, -1 };

struct node
{
    int x, y;
};
void solve()
{
    queue<node> q;
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < m; j++)
        {
            if (a[i][j] == '0')
            {
                vis[i][j] = true;
                q.push(node{i, j});
            }
        }
    }
    while (!q.empty())
    {
        node tmp = q.front();
        q.pop();
        for (int i = 0; i < 4; i++)
        {
            int nx = tmp.x + dx[i];
            int ny = tmp.y + dy[i];
            if (nx >= 0 && nx < n && ny >= 0 && ny < m && !vis[nx][ny] && a[nx][ny] == '1')
            {
                vis[nx][ny] = true;
                d[nx][ny] = d[tmp.x][tmp.y] + 1;
                q.push(node{nx, ny});
            }
        }
    }
}

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