下图给出了一个迷宫的平面图，其中标记为1 的为障碍，标记为0 的为可以通行的地方。
010000 000100 001001 110000 迷宫的入口为左上角，出口为右下角，在迷宫中，只能从一个位置走到这个它的上、下、左、右四个方向之一。 对于上面的迷宫，从入口开始，可以按DRRURRDDDR 的顺序通过迷宫，一共10 步。其中D、U、L、R 分别表示向下、向上、向左、向右走。对于下面这个更复杂的迷宫（30 行50 列），请找出一种通过迷宫的方式，其使用的步数最少，在步数最少的前提下，请找出字典序最小的一个作为答案。请注意在字典序中D<L<R<U。
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#include<iostream> #include<vector> #include<queue> using namespace std; struct node{ int x,y; vector<char> path; }; int dx[4] = {-1,0,1,0},dy[4] = {0,1,0,-1}; vector<char> ans(1500,0); vector<string> maze; int k = 0; void bfs(int x,int y){ node now,next; now.x = x,now.y = y; queue<node> q; q.push(now); maze[x][y] = '1'; while(q.size()){ now = q.front(); q.pop(); for(int i = 0;i < 4;i ++){ next.x = now.x + dx[i],next.y = now.y + dy[i],next.path = now.path; if(next.x >= 0 && next.x < 30 && next.y >= 0 && next.y < 50 && maze[next.x][next.y] == '0'){ if(i == 0) next.path.push_back('U'); else if(i == 1) next.path.push_back('R'); else if(i == 2) next.path.push_back('D'); else next.path.push_back('L'); q.push(next); maze[next.x][next.y] = '1'; if(next.x == 29 && next.y == 49){ if(next.path.size() < ans.size() || next.path.size() == ans.size() && next.path < ans){ ans = next.path; } } } } } } int main() { string s; for(int i = 0;i < 30;i ++){ cin>>s; maze.push_back(s); } bfs(0,0); for(int i = 0;i < ans.size();i ++){ cout<<ans[i]; } cout<<endl; return 0; }
