#include <iostream>
using namespace std;
char a[150][150];
int r[150][3];
int n;
void show(int k)
{
    for(int i = 1;i < k;i++)
    {
        cout<<"("<<r[i][1]<<","<<r[i][2]<<")->";
    }
     
    cout<<"("<<r[k][1]<<","<<r[k][2]<<")"<<endl;
}
//从x、y点开始逐步探测 
void num(int x,int y,int k)
{
	r[k][1] = x;
    r[k][2] = y;
    a[x][y] = '1';//探测过的点，设置为不可探测，防止重复的探测 
    if(x == n && y == n)
    {
        show(k);
    }
    else
    {
        //左、上、右、下探测，且不越界 
        if(y - 1 >= 1 && a[x][y - 1] == '0') num(x,y - 1,k+1);
        if(x - 1 >= 1 && a[x - 1][y] == '0') num(x-1,y,k+1);
        if(y + 1 <= n && a[x][y + 1] == '0') num(x,y + 1,k+1);
        if(x + 1 <= n && a[x + 1][y] == '0') num(x+1,y,k+1);
    }
} 
int main()
{
    int i,j;
    cin>>n;
    //0表示能走，1表示不能走 
    for(i = 1;i <= n;i++)
    {
        for(j = 1;j <= n;j++)
        {
            cin>>a[i][j];
        }
    }
    num(1,1,1);
    return 0;
}//已AC

这道题用深搜dfs

#include<bits/stdc++.h>
using namespace std;
bool f;
char c[30][30];
int n,a1[407],a2[407],dx[]={0,-1,0,1},dy[]={-1,0,1,0};
bool flag;
void re(int d1)
{
    flag=1;
    for (int i=1;i<d1;i++)
        cout<<"("<<a1[i]<<","<<a2[i]<<")->";
    cout<<"("<<a1[d1]<<","<<a2[d1]<<")";
    return;
}
void dfs(int x,int y,int d)
{
    if (flag)
        return;
    a1[d]=x,a2[d]=y;
    c[x][y]+=1;
    if (x==n&&y==n)
        re(d);
    for (int i=0;i<4;i++)
        if (x+dx[i]>0&&x+dx[i]<=n&&y+dy[i]>0&&y+dy[i]<=n&&c[x+dx[i]][y+dy[i]]=='0')
            dfs(x+dx[i],y+dy[i],d+1);
    return;
}
int main()
{
    cin>>n;
    for (int i=1;i<=n;i++)
    {
        string s;
        cin>>s;
        for (int j=1;j<=n;j++)
            c[i][j]=s[j-1];
    }
    dfs(1,1,1);
    return 0;
}

这道题是一道典型深搜，让我们输出路线。难点在于存储走过的点。 解决方案：用二维数组b存储走过的x,y点，随后输出。

#include <bits/stdc++.h>//b数组可以用结构体代替
using namespace std;
int n,b[400][10];
char a[30][30];
//方向的变化———左，上，右，下
int fx[5]={0,0,-1,0,1};
int fy[5]={0,-1,0,1,0};
void print(int s){//打印第一条路线
    for (int i=1;i<=s;i++){
        cout<<'('<<b[i][1]<<','<<b[i][2]<<")";
        if (i!=s)cout<<"->";//不是最后一个便在后面加上->
    }
    exit(0);//停止程序
}
void fun(int x,int y,int k){//递归搜索
    //存储b数组，标记a数组
    a[x][y]='1';
    b[k][1]=x;
    b[k][2]=y;
    if (x==n&&y==n)print(k);//判断是否到了终点，到了终点便执行函数
    //循环方向
    for (int i=1;i<=4;i++){
        //求出当前点位在左，上，右，下后分别在哪
        int tx=x+fx[i];
        int ty=y+fy[i];
        if (a[tx][ty]=='0')fun(tx,ty,k+1);//判断（是否在迷宫内，是否被标记过）并递归求解
    }
}
int main(){
    //输入
    cin>>n;
    for (int i=1;i<=n;i++){
        for (int j=1;j<=n;j++)cin>>a[i][j];
    }
    //调用函数
    fun(1,1,1);
    return 0;
}

我用了dfs

#include <bits/stdc++.h>
using namespace std;
char x;
int n,m,a[105][105],used[105][105],minn = 1e9,num = 1;
struct Point
{
	int xx,yy;
} w[100005]; //w[i]表示第i步走到的位置，一定要定大一点，因为不知道要走多少步
void f()
{
	cout << "(1,1)";
	for (int i = 1;i <= num;i++)
	{
		cout << "->(" << w[i].xx << "," << w[i].yy << ")";
	}
}
void dfs(int x,int y,int ans)
{
	if (a[x][y] == 1) return;
	if (x > n || y > n) return;
	if (x < 1 || y < 1) return;
	w[ans] = (Point){x,y};
	if (x == n && y == n)
	{
		num = ans;
		f();
		exit(0);
	}
	if (used[x][y] == 0)
	{
		used[x][y] = 1;
		dfs(x,y-1,ans+1);     //顺序一定不能错！我调了好久才调对~(左上右下)
		dfs(x-1,y,ans+1);
		dfs(x,y+1,ans+1);
		dfs(x+1,y,ans+1);
		used[x][y] = 0;
	}
}
int main()
{
	cin >> n;
	for (int i = 1;i <= n;i++)
	{
		for (int j = 1;j <= n;j++)
		{
			cin >> x;
			if (x != '1' && x != '0')  //第49~53行可以不要，这是防止输入错误。机器自动输入应该不会输错
			{
				j--;
				continue;
			}
			a[i][j] = x-'0';
		}
	}
	dfs(1,1,0);
	return 0;
}
/*
这是我的自制迷宫！！没事可以走走看！
	21 24
	0,0,1,0,0,0,1,0,0,1,0,0,1,0,0,0,1,0,0,0,0,0,0,1
	1,0,1,0,1,0,1,0,0,0,1,0,1,0,1,0,0,0,1,0,1,1,0,1
	1,0,0,0,1,0,1,0,1,0,1,0,1,0,1,0,0,1,1,0,0,1,0,1
	1,0,1,1,1,0,0,0,1,0,1,0,1,0,1,0,0,1,0,0,0,1,0,1
	1,0,0,0,0,1,1,1,0,0,0,0,1,0,1,0,0,0,0,1,0,1,0,1
	1,0,1,1,0,0,1,0,0,1,1,0,1,0,1,0,1,0,1,1,0,0,0,0
	1,0,0,0,1,0,1,1,1,0,0,0,1,0,1,0,1,0,1,1,0,0,1,0
	1,0,1,0,1,0,1,1,1,0,0,0,1,0,1,0,1,0,1,1,1,0,0,1
	1,0,0,0,1,0,0,1,0,0,1,0,1,0,1,0,1,0,0,0,0,1,1,1
	1,1,0,1,0,0,1,0,0,1,0,0,1,0,1,0,1,0,1,1,1,0,0,0
	1,0,0,1,0,1,1,0,1,0,1,1,1,0,0,0,1,0,1,1,1,0,0,0
	1,0,1,0,0,1,0,0,1,0,1,0,1,0,1,1,0,0,0,0,0,0,1,1
	1,0,1,0,1,1,0,1,0,1,0,0,1,0,0,0,0,1,0,1,1,0,1,0
	1,0,0,0,1,1,0,1,0,1,1,1,1,0,1,1,1,0,0,1,0,0,0,0
	1,1,1,0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,1,0,0,1,1,0
	0,0,1,0,0,0,1,0,1,1,1,1,1,0,0,0,1,0,1,0,0,0,0,0
	0,0,1,1,0,0,1,0,1,0,0,1,0,0,1,1,1,0,0,1,0,1,1,0
	1,0,0,0,0,0,0,1,0,1,0,1,0,1,1,0,0,1,0,0,0,0,0,0
	1,0,1,0,0,1,0,1,0,1,0,1,0,1,0,0,0,1,0,1,1,1,1,1
	1,0,0,1,0,1,0,1,0,1,0,1,0,1,1,0,0,1,0,1,1,0,0,0
	1,1,0,0,1,1,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,0,1,0
*/

希望大家能点赞！

大家能帮我看看哪里错了吗？33分（谢谢啦！）

#include<bits/stdc++.h>
using namespace std;
int n;
char a[25][25];
struct Point
{
	int x,y;
} d[25];  //存储迷宫路线
int dx[4] = {-1,0,1,0},dy[4] = {0,1,0,-1};   //方向数组（左上右下）
bool used[25][25];
void dfs(int x,int y,int num)   //num表示走了几步
{
	used[x][y] = 1;
	if (x == n && y == n)
	{
		printf("(1,1)");
		for (int i = 1;i < num;i++)
		{
			printf("->(%d,%d)",d[i].x,d[i].y);
		}
		exit(0);
	}
	for (int i = 0;i < 4;i++)
	{
		int nx = x + dx[i],ny = y + dy[i];
		if (nx < 1 || ny < 1 || nx > n || ny > n || a[nx][ny] == '1' || used[nx][ny] == 1)
			continue;
		d[num] = ((Point){nx,ny});
		dfs(nx,ny,num+1);
	}
	used[x][y] = 0;
}
int main()
{
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	
	cin >> n;
	for (int i = 1;i <= n;i++)
	{
		for (int j = 1;j <= n;j++)
		{
			cin >> a[i][j];
		}
	}
	dfs(1,1,1);
	return 0;
}

写题解请注意

鼓励大家写题解，但注意题解格式。

题解一定要有思路解析或代码注释，能否让别人理解你的思路

也是你的能力的检验，不要只放无意义的代码给大家复制，那就失去了做题的初心。

给代码两端加上这个会舒服一些

```cpp

你的代码

```

</span>

这个点在键盘的左上角tab上面那个键，注意切换输入法

#include<iostream>
using namespace std;
int main()
{
    int n;
    cin>>n;//这是一个注释
    return 0;
} 

请注意严禁抄袭题解，写题解不要只放代码，需加上你的思路或代码注释。

抄袭题解一经发现直接取消成绩。

题解被删除的可能

1. 代码不符合格式规范
2. 没有思路讲解或者没有注释，
3. 无意义的题解

大家携手共同维护一个良好的编程环境，如果一经发现，多次作乱。可能会被管理员拉黑，请注意，一旦拉黑即失去登陆资格。