怎么求多源点到多目的地的最短距离
原理是，设计一个超级远点S，S和每个源点建立一条边权为0的路径。（这个S是假想的，实际实现的时候可不需建立）
另一个原理是，队列的两个性质：1. 两端性，2.单调性
队列里面只可能出现 [x,x,x ..., x+1, x+1, ...]
不会同时出现x, x+1, x+2

实现的时候只需要把所有源点push到队列里就行了。

AcWing 173. 矩阵距离

#include<bits/stdc++.h>

using namespace std;

const int N = 1010;

char g[N][N];
int dist[N][N];
int n, m;

int main() {
    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
        scanf("%s", g[i] + 1);
    }

    queue<pair<int, int>> q;
    memset(dist, -1, sizeof dist);

    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            if (g[i][j] == '1') {
                dist[i][j] = 0;
                q.push({i, j});
            }
        }
    }

    int dx[] = {-1, 0, 1, 0}, dy[] = {0, 1, 0, -1};
    while (q.size()) {
        auto pos = q.front();
        q.pop();
        int x = pos.first, y = pos.second;
        for (int i = 0; i < 4; i++) {
            int a = x + dx[i], b = y + dy[i];
            if (a >= 1 && a <= n && b >= 1 && b <= m && dist[a][b] == -1) {
                dist[a][b] = dist[x][y] + 1;
                q.push({a, b});
            }
        }
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            cout << dist[i][j]<<" ";
        }
        cout << endl;
    }
}

AcWing 3205. 最优配餐

#include<bits/stdc++.h>

using namespace std;

const int N = 1010;

bool g[N][N];
int dist[N][N];
int n, m, k, d;

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

struct Tar {
    int x, y, c;
} trg[N * N];
queue<pair<int, int>> q;

void bfs() {
    while (q.size()) {
        auto u = q.front();
        q.pop();
        int x = u.first, y = u.second;
        for (int i = 0; i < 4; i++) {
            int a = x + dx[i], b = y + dy[i];
            if (a >= 1 && a <= n && b >= 1 && b <= n && !g[a][b]) {
                if (dist[a][b] > dist[x][y] + 1) {
                    dist[a][b] = dist[x][y] + 1;
                    q.push({a, b});
                }
            }
        }
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin >> n >> m >> k >> d;

    memset(dist, 0x3f, sizeof dist);

    while (m--) {
        int x, y;
        cin >> x >> y;
        q.push({x, y});
        dist[x][y] = 0;
    }
    for (int i = 0; i < k; i++) {
        int x, y, c;
        cin >> x >> y >> c;
        trg[i] = {x, y, c};
    }

    while (d--) {
        int x, y;
        cin >> x >> y;
        g[x][y] = true;
    }

    bfs();

    long long res = 0;
    for (int i = 0; i < k; i++) {
        int cost = 1ll * dist[trg[i].x][trg[i].y] * trg[i].c;
        res += cost;
    }
    cout << res << endl;
}