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 = 500+1;



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 = 0, int Y = 0){

        x = X; y = Y;

    }

};



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

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

//1,1

void rbfs(int id)

{

    char (*G)[maxn];

    int (*vis)[maxn];

    if(id == 1) G = g1, vis = d1;

    else G = g2, vis = d2;

    memset(vis,-1,sizeof(d1));

    queue<node>q;

    node u(n-1,m-1);

    q.push(u);

    vis[u.x][u.y] = 0;

    while(q.size()){

        u = q.front();q.pop();

        if(u.x == 0 && u.y == 0) return;

        for(int i = 0; i < 4; i++){

            int nx = u.x + dx[i], ny = u.y + dy[i];

            if(nx>=0&&nx<n&&ny>=0&&ny<m&&G[nx][ny]!='#'&&!~vis[nx][ny]){

                vis[nx][ny] = vis[u.x][u.y]+1;

                q.push(node(nx,ny));

            }

        }

    }

}

bool vis[maxn][maxn];



bool bfs()

{

    if(d1[0][0] != d2[0][0] )return false;

    memset(vis,0,sizeof(vis));

    queue<node>q;

    q.push(node(0,0));

    int tx = n-1, ty = m-1;

    while(q.size()){

        node &u = q.front();

        if(u.x == tx && u.y == ty) return true;

        for(int i = 0; i < 4; i++){

            int nx = u.x + dx[i], ny = u.y + dy[i];

            if(nx>=0&&nx<n&&ny>=0&&ny<m&&

               d1[nx][ny] == d1[u.x][u.y] - 1 && d2[nx][ny] == d2[u.x][u.y] - 1 && !vis[nx][ny]){

                vis[nx][ny] = 1;

                q.push(node(nx,ny));

            }

        }

        q.pop();

    }

    return false;

}



int main()

{

   // freopen("in.txt","r",stdin);

    scanf("%d%d",&n,&m);

    for(int i = 0; i < n; i++)

        scanf("%s",g1[i]);

     for(int i = 0; i < n; i++)

        scanf("%s",g2[i]);

    rbfs(1);

    rbfs(2);

    printf("%s\n",bfs()?"YES":"NO");

    return 0;

}