题解：

以3个大陆为起点，都dfs一遍，求出该大陆到其他点的最小距离是多少， 然后枚举每个点作为3个大陆的路径交点。

代码：

#include
      
       using namespace std;#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);#define LL long long#define ULL unsigned LL#define fi first#define se second#define pb push_back#define ep emplace_back#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1#define lch(x) tr[x].son[0]#define rch(x) tr[x].son[1]#define max3(a,b,c) max(a,max(b,c))#define min3(a,b,c) min(a,min(b,c))typedef pair
       
         pll;const int inf = 0x3f3f3f3f;const int _inf = 0xc0c0c0c0;const LL INF = 0x3f3f3f3f3f3f3f3f;const LL _INF = 0xc0c0c0c0c0c0c0c0;const LL mod =  (int)1e9+7;const int N = 1e3 + 100;int n, m;LL dis[N][N][3];char s[N][N];int dx[] = {
    1, -1, 0, 0};int dy[] = {
    0, 0, -1, 1};deque
        
          q;void bfs(int k){    for(int i = 1; i <= n; ++i)        for(int j = 1; j <= m; ++j)            dis[i][j][k] = inf;    char kk = k + '0' + 1;    for(int i = 1; i <= n; ++i){        for(int j = 1; j <= m; ++j){            if(s[i][j] == kk){                dis[i][j][k] = 0;                q.push_back(pll(i,j));            }        }    }    while(!q.empty()){        pll t = q.front();        q.pop_front();        for(int _ = 0; _ < 4; ++_){            int nx = t.fi + dx[_];            int ny = t.se + dy[_];            if(nx < 1 || nx > n || ny < 1 || ny > m || s[nx][ny] == '#') continue;            int dd = dis[t.fi][t.se][k] + (s[nx][ny] == '.');            if(dd < dis[nx][ny][k]){                dis[nx][ny][k] = dd;                if(s[nx][ny] == '.') q.push_back(pll(nx,ny));                else q.push_front(pll(nx, ny));            }        }    }}int main(){    scanf("%d%d", &n, &m);    for(int i = 1; i <= n; ++i)        scanf("%s", s[i]+1);    for(int i = 0; i < 3; ++i) bfs(i);    LL ans = inf;    for(int i = 1; i <= n; ++i){        for(int j = 1; j <= m; ++j){            int f = 0;            if(s[i][j] == '.') f = 2;            ans = min(ans, dis[i][j][0] + dis[i][j][1] + dis[i][j][2] - f);        }    }    if(ans == inf) ans = -1;    cout << ans << endl;    return 0;}