2021.11.04 bfs
题目名：矩阵距离
题目链接：https://www.acwing.com/problem/content/175/
思路：把所有的”1”入队，然后bfs就ok了

#include<bits/stdc++.h>
using namespace std;
using pii = pair<int, int>;
const int N = 1010;
int a[N][N], dist[N][N];
int dx[4] = {1,-1,0,0}, dy[4] = {0,0,1,-1};
int n, m;

queue<pii> q;

bool value(int x, int y){
    return ~x && ~y && x < n && y < m;
}

int main(){
    cin >> n >> m;
    memset(dist, 0x3f, sizeof dist);
    for(int i = 0; i < n; i++){
        string s; cin >> s;
        for(int j = 0; j < m; j++)
            if(s[j] == '1') a[i][j] = 1, q.push({i,j}), dist[i][j] = 0;

    }

    while(q.size()){
        int x = q.front().first, y = q.front().second;
        q.pop();
        for(int i = 0; i < 4; i++){
            int nx = x + dx[i], ny = y + dy[i];
            if(value(nx, ny) && dist[nx][ny] == 0x3f3f3f3f){
                q.push({nx, ny});
                dist[nx][ny] = dist[x][y] + 1;
            }
        }
    }

    for(int i = 0; i < n; i++){
        for(int j = 0; j < m; j++)
            cout << dist[i][j] << ' ';
        cout << "\n";    
    }

    return 0;
}