1.直接对bfs，记录f[i][j]：A到(i, j)的最小时间。然后B在bfs时就可以统计答案了。
2.当然可以双向bfs。

```1.
#include <bits/stdc++.h>
#define LL long long
using namespace std;

char a[1005][1005];
int xx[]={0, 0, 1, -1, -1, -1, 1, 1};
int yy[]={1, -1, 0, 0, 1, -1, -1, 1};
struct node{
int x, y, t;
};
queue<node> q;
int vis[1005][1005], f[1005][1005], ans=1<<30;
int n, m;

void BFS1(int x, int y){
memset(vis, 0, sizeof(vis));
memset(f, -1, sizeof(f));
q.push(node{x, y, 0});
vis[x][y]=1;
while(!q.empty()){
node now=q.front(); q.pop();
f[now.x][now.y]=now.t;
for(int i=0; i<8; i++){
int x=now.x+xx[i], y=now.y+yy[i];
if(!vis[x][y]&&a[x][y]!='#'&&x>=1&&x<=n&&y>=1&&y<=m){
q.push(node{x, y, now.t+1}); vis[x][y]=1;
}
}
}
}

void BFS2(int x, int y){
memset(vis, 0, sizeof(vis));
q.push(node{x, y, 0});
vis[x][y]=1;
while(!q.empty()){
node now=q.front(); q.pop();
if(f[now.x][now.y]!=-1){
ans=min(ans, max((now.t+1)/2, f[now.x][now.y]));
}

for(int i=0; i<4; i++){
int x=now.x+xx[i], y=now.y+yy[i];
if(!vis[x][y]&&a[x][y]!='#'&&x>=1&&x<=n&&y>=1&&y<=m){
q.push(node{x, y, now.t+1}); vis[x][y]=1;
}
}
}
}

int main(){
scanf("%d%d%*c", &n, &m);
int sx, sy, tx, ty;
for(int i=1; i<=n; i++){
for(int j=1; j<=m; j++){
scanf("%c%*c", &a[i][j]);
if(a[i][j]=='C'){
sx=i, sy=j;
}
if(a[i][j]=='D'){
tx=i, ty=j;
}
}
}
BFS1(sx, sy);
BFS2(tx, ty);
if(ans==1<<30){
printf("NO\n");
}
else{
printf("YES\n");
printf("%d\n", ans);
}

return 0;
}
*************************************
2.
#include <bits/stdc++.h>
#define LL long long
using namespace std;

char a[1005][1005];
int xx[]={0, 0, 1, -1, -1, -1, 1, 1};
int yy[]={1, -1, 0, 0, 1, -1, -1, 1};
struct node{
int x, y;
};
queue<node> q[2];
int vis[2][1005][1005];
int n, m;

int bfs(int pos){
int siz=q[pos].size();
while(siz--){
node now=q[pos].front(); q[pos].pop();
for(int k=0; k<4+(pos?0:4); k++){
int x=now.x+xx[k], y=now.y+yy[k];
if(x>=1&&x<=n&&y>=1&&y<=m&&!vis[pos][x][y]&&a[x][y]!='#'){
if(vis[!pos][x][y]){
return 1;
}
vis[pos][x][y]=1;
q[pos].push(node{x, y});
}
}
}
return 0;
}

int main(){
scanf("%d%d%*c", &n, &m);
int sx, sy, tx, ty;
for(int i=1; i<=n; i++){
for(int j=1; j<=m; j++){
scanf("%c%*c", &a[i][j]);
if(a[i][j]=='C'){
sx=i, sy=j;
}
if(a[i][j]=='D'){
tx=i, ty=j;
}
}
}
int ans=0;
q[0].push(node{sx, sy}); vis[0][sx][sy]=1;
q[1].push(node{tx, ty}); vis[1][tx][ty]=1;
while(!q[0].empty()||!q[1].empty()){
ans++;
if(bfs(0)||bfs(1)||bfs(1)){
printf("YES\n");
printf("%d\n", ans);
return 0;
}
}
printf("NO\n");

return 0;
}
```