走玉米地迷宫,一般有两种简单策略,遇到岔路总是优先沿着自己的左手方向,或者右手方向走。给一个迷宫,给出这两种策略的步数,再给出最短路径的长度。

#########
#.#.#.#.#
S.......E
#.#.#.#.#
######### 迷宫示例
前两种策略用DFS的方法,最后一种用BFS的方法分别确定路线。易错点:
1.这道题目因为在走的时候遇到死路需要回头,且这个步数也要计算在内,所以不需要将走过的路设置标志。
2.总是沿着当前的左边或者右边走,即方向需要不断调整
贴代码如下:
 #include<iostream>
#include<string>
#include<queue>
using namespace std;
typedef class
{
public:
int x,y;
int path;
}vertex; char maze[][];
int trace[][];
int lpath,rpath,spath;
queue<vertex> bfq; int incre[][]={ //first LEFT, then MID, last RIGHT
{-,,,,,,,-},
{,,,,,-,-,},
{,,,-,-,,,},
{,-,-,,,,,}
}; int LDFS(int x,int y,int ind)
{
int nind;
trace[x][y]=;
if(maze[x][y]=='E')
{
cout<<lpath<<" ";
return -;
}
//左转
if(maze[x+incre[ind][]][y+incre[ind][]]!='#'&&trace[x+incre[ind][]][y+incre[ind][]]==)
{
lpath++;
nind=(ind+)%;
if(LDFS(x+incre[ind][],y+incre[ind][],nind)==-) return -;
}
//直行
if(maze[x+incre[ind][]][y+incre[ind][]]!='#'&&trace[x+incre[ind][]][y+incre[ind][]]==)
{
lpath++;
if(LDFS(x+incre[ind][],y+incre[ind][],ind)==-) return -;
}
//右转
if(maze[x+incre[ind][]][y+incre[ind][]]!='#'&&trace[x+incre[ind][]][y+incre[ind][]]==)
{
lpath++;
nind=(ind+)%;
if(LDFS(x+incre[ind][],y+incre[ind][],nind)==-) return -;
}
//后转
if(maze[x+incre[ind][]][y+incre[ind][]]!='#'&&trace[x+incre[ind][]][y+incre[ind][]]==)
{
lpath++;
nind=(ind+)%;
if(LDFS(x+incre[ind][],y+incre[ind][],nind)==-) return -;
} return lpath;
} int RDFS(int x,int y,int ind)
{
int nind;
trace[x][y]=;
if(maze[x][y]=='E')
{
cout<<rpath<<" ";
return -;
}
//右转
if(maze[x+incre[ind][]][y+incre[ind][]]!='#'&&trace[x+incre[ind][]][y+incre[ind][]]==)
{
rpath++;
nind=(ind+)%;
if(RDFS(x+incre[ind][],y+incre[ind][],nind)==-) return -;
}
if(maze[x+incre[ind][]][y+incre[ind][]]!='#'&&trace[x+incre[ind][]][y+incre[ind][]]==)
{
rpath++; if(RDFS(x+incre[ind][],y+incre[ind][],ind)==-) return -;
}
if(maze[x+incre[ind][]][y+incre[ind][]]!='#'&&trace[x+incre[ind][]][y+incre[ind][]]==)
{
rpath++;
nind=(ind+)%;
if(RDFS(x+incre[ind][],y+incre[ind][],nind)==-) return -;
}
if(maze[x+incre[ind][]][y+incre[ind][]]!='#'&&trace[x+incre[ind][]][y+incre[ind][]]==)
{
rpath++;
nind=(ind+)%;
if(RDFS(x+incre[ind][],y+incre[ind][],nind)==-) return -;
}
return rpath;
} void BFS(vertex v,int ind)
{
vertex a,b;
bfq.push(v);
trace[v.x][v.y]=;
while(bfq.size()!=)
{
a=bfq.front();
bfq.pop();
if(maze[a.x+incre[ind][]][a.y+incre[ind][]]!='#'&&trace[a.x+incre[ind][]][a.y+incre[ind][]]==)
{
trace[a.x+incre[ind][]][a.y+incre[ind][]]=;
b.x=a.x+incre[ind][]; b.y=a.y+incre[ind][]; b.path=a.path+;
if(maze[b.x][b.y]=='E')
{
cout<<b.path<<endl;
break;
}
bfq.push(b);
}
if(maze[a.x+incre[ind][]][a.y+incre[ind][]]!='#'&&trace[a.x+incre[ind][]][a.y+incre[ind][]]==)
{
trace[a.x+incre[ind][]][a.y+incre[ind][]]=;
b.x=a.x+incre[ind][]; b.y=a.y+incre[ind][]; b.path=a.path+;
if(maze[b.x][b.y]=='E')
{
cout<<b.path<<endl;
break;
}
bfq.push(b);
}
if(maze[a.x+incre[ind][]][a.y+incre[ind][]]!='#'&&trace[a.x+incre[ind][]][a.y+incre[ind][]]==)
{
trace[a.x+incre[ind][]][a.y+incre[ind][]]=;
b.x=a.x+incre[ind][]; b.y=a.y+incre[ind][]; b.path=a.path+;
if(maze[b.x][b.y]=='E')
{
cout<<b.path<<endl;
break;
}
bfq.push(b);
}
if(maze[a.x+incre[ind][]][a.y+incre[ind][]]!='#'&&trace[a.x+incre[ind][]][a.y+incre[ind][]]==)
{
trace[a.x+incre[ind][]][a.y+incre[ind][]]=;
b.x=a.x+incre[ind][]; b.y=a.y+incre[ind][]; b.path=a.path+;
if(maze[b.x][b.y]=='E')
{
cout<<b.path<<endl;
break;
}
bfq.push(b);
}
}
} int main()
{
int instan,w,h,i,j,sx,sy,ind;
char in;
cin>>instan;
for(int Num=;Num<instan;Num++)
{
cin>>w>>h;
for(i=;i<h;i++)
for(j=;j<w;j++)
{
cin>>in;
maze[i][j]=in;
if(maze[i][j]=='S')
{
sx=i;sy=j;
}
}
if(sy==) ind=;
if(sx==) ind=;
if(sy==(w-)) ind=;
if(sx==(h-)) ind=;
lpath=;
memset(trace,,sizeof(trace));
lpath=LDFS(sx,sy,ind);
rpath=,spath=;
memset(trace,,sizeof(trace));
rpath=RDFS(sx,sy,ind);
//cout<<lpath<<" "<<rpath<<endl;
vertex vs;
vs.x=sx;vs.y=sy;vs.path=;
memset(trace,,sizeof(trace));
BFS(vs,ind);
while(bfq.empty()!=)
{
bfq.pop();
}
}
return ;
}
05-04 04:11