题解 | #迷宫问题#dfs解法 ！可以根据输入参数step和vector得出当路径不唯一时输出最短路径

#include<iostream>
#include<vector>
//#include<algorithm>
#include<cstdio>
using namespace std;
int a[100][100] ;
int v[100][100] = {0};
int minx = 999;
int n=0,m=0;
int x,y,p,q;
int nx =0,ny=0;
vector<int> xx,yy;
void dfs(int x,int y,int step)
{
	if(x == p && y == q)	
	{
        for(int i=0;i<yy.size();i++)
        cout<<'('<<xx[i]<<','<<yy[i]<<')'<<endl;
		return;	
	}
	if(a[x][y+1] == 0 && v[x][y+1] == 0)
	{
		v[x][y+1] = 1;
        ny++;
        xx.push_back(nx);
        yy.push_back(ny);
		dfs(x,y+1,step+1);
        xx.pop_back();
        yy.pop_back();
        ny--;
		v[x][y+1] = 0;
	}
		if(a[x+1][y] == 0 && v[x+1][y] == 0)
	{
		v[x+1][y] = 1;
        nx +=1;
        xx.push_back(nx);
        yy.push_back(ny);
		dfs(x+1,y,step+1);
        xx.pop_back();
        yy.pop_back();
            nx--;
		v[x+1][y] = 0;
	}
		if(a[x][y-1] == 0 && v[x][y-1] == 0)
	{
		v[x][y-1] = 1;
        ny-=1;
        xx.push_back(nx);
        yy.push_back(ny);
		dfs(x,y-1,step+1);
        xx.pop_back();
        yy.pop_back();
            ny++;
		v[x][y-1] = 0;
	}	
        if(a[x-1][y] == 0 && v[x-1][y] == 0)
	{
        v[x-1][y] = 1;
        nx -=1;
        xx.push_back(nx);
        yy.push_back(ny);
		dfs(x-1,y,step+1);
        xx.pop_back();
        yy.pop_back();
            nx++;
		v[x-1][y] = 0;
	}
	return;
 } 
int main(void)
{
    scanf("%d%d",&m,&n);
    for(int i=0;i<=99;i++)
	{
		for(int j=0;j<=99;j++)
		{
			a[i][j] =3;
		}
	}
	int i=1,j=1;
	
	for(;i<=m;i++)
	{
		for(j=1;j<=n;j++)
		{
			scanf("%d",&a[i][j]);
		}
	}
	p=m;
    q=n;
    x=1;
    y=1;
    nx=0;ny=0;
	v[x][y] = 1;
    cout<<"(0,0)"<<endl;
	dfs(x,y,0);
	return 0;
}