[AcWing 173] 矩阵距离

多源 BFS

---

点击查看代码

#include<bits/stdc++.h>

using namespace std;

typedef long long LL;

const int N = 1000 + 10;

#define x first
#define y second

int n, m;
char g[N][N];
int d[N][N];
int dx[] = {-1, 0, 1, 0};
int dy[] = {0, 1, 0, -1};

void bfs()
{
    memset(d, -1, sizeof d);
    queue<pair<int,int>> q;
    for (int i = 1; i <= n; i ++)
        for (int j = 1; j <= m; j ++)
            if (g[i][j] == '1') {
                q.push({i, j});
                d[i][j] = 0;
            }
    while (q.size()) {
        auto t = q.front();
        q.pop();
        for (int i = 0; i < 4; i ++) {
            int a = t.x + dx[i], b = t.y + dy[i];
            if (a < 1 || a > n || b < 1 || b > m || d[a][b] != -1)
                continue;
            d[a][b] = d[t.x][t.y] + 1;
            q.push({a, b});
        }
    }
}

void solve()
{
    cin >> n >> m;
    for (int i = 1; i <= n; i ++)
        for (int j = 1; j <= m; j ++)
            cin >> g[i][j];
    bfs();
    for (int i = 1; i <= n; i ++) {
        for (int j = 1; j <= m; j ++)
            cout << d[i][j] << ' ';
        cout << endl;
    }
}

signed main()
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);

    solve();

    return 0;
}

---

1. 可以假象一个单源点，源点到所有为 "1" 的点的距离为 $0$，之后就转化为单源最短路模型，用 $BFS$ 求解