多源bfs

多源bfs

1.算法分析

多源bfs就是一开始有很多的源头，不只一个源点。处理的方法:一开始直接把所有源点放入队列；或者建立一个虚拟源点，虚拟源点到每个源点的距离为0

2.例题

acwing173矩阵距离
给定一个N行M列的01矩阵A，A[i][j] 与 A[k][l] 之间的曼哈顿距离定义为：
dist(A[i][j],A[k][l])=|i−k|+|j−l|
输出一个N行M列的整数矩阵B，其中：
B[i][j]=min(1≤x≤N,1≤y≤M,A[x][y]=1)dist(A[i][j],A[x][y])
N、M~1e3

#include 

using namespace std;

struct POS {
    int x, y, step;
};
int const N = 1e3 + 10;
int a[N][N], b[N][N];
int n, m;
queue q;

int dx[] = {1, 0, -1, 0}, dy[] = {0, 1, 0, -1};

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

int main() {
    memset(b, -1, sizeof b);
    cin >> n >> m;
    for (int i = 1; i <= n; ++i) {
        string s;
        cin >> s;
        for (int j = 0; j < s.size(); ++j) {
            a[i][j + 1] = s[j] - '0';
            if (a[i][j + 1]) {
                b[i][j + 1] = 0;
                q.push({i, j + 1, 0});  // 把所有源点放入队列
            }
        }
    }
    
    bfs();
    
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) 
            cout << b[i][j] << " ";
        cout << endl;
    }
    return 0;
}