poj3083——Children of the Candy Corn

题意：分别找出从S到E的左路优先、右路优先、最短路的step。

思路：dfs+bfs.

#include<iostream>#include<cstdio>#include<string>#include<queue>using namespace std;#define N 45char maze[N][N];bool vis[N][N],flag;int n,m,leftstep,rightstep;int leftdir[][2]={{0,-1},{-1,0},{0,1},{1,0}};int rightdir[][2]={{0,1},{-1,0},{0,-1},{1,0}};class point{public:int x,y;int step;};bool boundary(point a){if(a.x >=0&&a.x<n&&a.y >=0&&a.y <m)return true;return false;}int shortest(point st,point ed){int i,j,k;int dir[][2]={{0,-1},{0,1},{-1,0},{1,0}};queue<point >q;memset(vis,true,sizeof(vis));point node,p;node=st;node.step =1;vis[node.x ][node.y ]=false;q.push (node);while(!q.empty()){node=q.front ();q.pop ();for(i=0;i<4;i++){p.x =node.x +dir[i][0];p.y =node.y +dir[i][1];if(boundary(p)&&vis[p.x ][p.y ]&&maze[p.x ][p.y ]!='#'){vis[p.x ][p.y ]=false;p.step =node.step +1;if(p.x ==ed.x &&p.y ==ed.y )return p.step ;q.push (p);}}}}void leftdfs(point st,point ed,int di,int step){point node,p;int i;if(flag) return ;if(st.x ==ed.x &&st.y==ed.y ){flag=true;leftstep=step;return ;}node=st;for(i=0;i<4;i++){p.x =node.x +leftdir[(di+3+i)%4][0];p.y =node.y +leftdir[(di+3+i)%4][1];if(flag ) return ;if(boundary(p)&&maze[p.x ][p.y ]!='#')leftdfs(p,ed,(di+3+i)%4,step+1);}}void rightdfs(point st,point ed,int di,int step){point node,p;int i;if(flag) return ;if(st.x==ed.x&&st.y ==ed.y ){flag=true;rightstep=step;return ;}node=st;for(i=0;i<4;i++){p.x =node.x+rightdir[(di+3+i)%4][0];p.y =node.y+rightdir[(di+3+i)%4][1];if(flag) return;if(boundary(p)&&maze[p.x ][p.y ]!='#')rightdfs(p,ed,(di+3+i)%4,step+1);}}int main(){int t,i,j;point st,ed;cin>>t;while(t--){cin>>m>>n;for(i=0;i<n;i++)scanf("%s",&maze[i]);for(i=0;i<n;i++)for(j=0;j<m;j++){if(maze[i][j]=='S') {st.x =i;st.y =j;}else if(maze[i][j]=='E'){ed.x =i;ed.y =j;}}point node=st,p;leftstep=0;rightstep=0;for(i=0;i<4;i++){p.x=node.x +leftdir[i][0];p.y=node.y +leftdir[i][1];if(boundary(p)&&maze[p.x ][p.y ]!='#'){flag=false;leftdfs(st,ed,i,1);break;}}for(i=0;i<4;i++){p.x =node.x +rightdir[i][0];p.y =node.y +rightdir[i][1];if(boundary(p)&&maze[p.x ][p.y ]!='#'){flag=false;rightdfs(st,ed,i,1);break;}}cout<<leftstep<<" "<<rightstep<<" ";cout<<shortest(st,ed)<<endl;}return 0;}