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;
}

转载于:https://www.cnblogs.com/jerryRey/p/4658378.html

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