CF Gym 100187E Two Labyrinths （迷宫问题）

题意：问两个迷宫是否存在公共最短路。

题解：两个反向bfs建立层次图，一遍正向bfs寻找公共最短路

#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int maxn = +;

int d1[maxn][maxn];
int d2[maxn][maxn];

char g1[maxn][maxn];
char g2[maxn][maxn];

int n,m;
struct node{
    int x,y;
    node(int X = , int Y = ){
        x = X; y = Y;
    }
};

int dx[] = {,-,,};
int dy[] = {,,,-};
//1,1
void rbfs(int id)
{
    char (*G)[maxn];
    int (*vis)[maxn];
    if(id == ) G = g1, vis = d1;
    else G = g2, vis = d2;
    memset(vis,-,sizeof(d1));
    queue<node>q;
    node u(n-,m-);
    q.push(u);
    vis[u.x][u.y] = ;
    while(q.size()){
        u = q.front();q.pop();
        if(u.x ==  && u.y == ) return;
        for(int i = ; i < ; i++){
            int nx = u.x + dx[i], ny = u.y + dy[i];
            if(nx>=&&nx<n&&ny>=&&ny<m&&G[nx][ny]!='#'&&!~vis[nx][ny]){
                vis[nx][ny] = vis[u.x][u.y]+;
                q.push(node(nx,ny));
            }
        }
    }
}
bool vis[maxn][maxn];

bool bfs()
{
    if(d1[][] != d2[][] )return false;
    memset(vis,,sizeof(vis));
    queue<node>q;
    q.push(node(,));
    int tx = n-, ty = m-;
    while(q.size()){
        node &u = q.front();
        if(u.x == tx && u.y == ty) return true;
        for(int i = ; i < ; i++){
            int nx = u.x + dx[i], ny = u.y + dy[i];
            if(nx>=&&nx<n&&ny>=&&ny<m&&
               d1[nx][ny] == d1[u.x][u.y] -  && d2[nx][ny] == d2[u.x][u.y] -  && !vis[nx][ny]){
                vis[nx][ny] = ;
                q.push(node(nx,ny));
            }
        }
        q.pop();
    }
    return false;
}

int main()
{
   // freopen("in.txt","r",stdin);
    scanf("%d%d",&n,&m);
    for(int i = ; i < n; i++)
        scanf("%s",g1[i]);
     for(int i = ; i < n; i++)
        scanf("%s",g2[i]);
    rbfs();
    rbfs();
    printf("%s\n",bfs()?"YES":"NO");
    return ;
}