60开不出来有问题，求调

__LTZ__ @ 2024-10-05 20:32:19

by __LTZ__ @ 2024-10-05 20:33:50

求求了，困扰本人3天了，国庆作业写不完了（嘤嘤嘤）

by _yAy_ @ 2024-10-05 20:41:42

能把代码粘出来吗

by Fractured_Angel @ 2024-10-05 20:45:40

@LTZ 不要把你写不完作业的原因归咎到调这么一道水题上吧。。。

by amd47802574 @ 2024-10-07 10:48:13

这题可以用洪水填充来解

//洪水填充
void f(int i,int j)
{
    if (i<1 || i>n || j<1 || j>m || c[i][j] != '.') return;
    c[i][j] = '0';// '.'-> '0'
    f(i - 1, j);
    f(i + 1, j);
    f(i, j - 1);
    f(i, j + 1);
}
f(1,1);//出发点

只要判断ch[n][m]是不是等于'0'就行了

by amd47802574 @ 2024-10-07 10:51:42

b站洪水填充讲解传送门

by Jchp @ 2024-10-10 20:09:13

x与y的顺序!!!

第一项是y第二项是x

#include <bits/stdc++.h>
using namespace std;

int n , m;
int maxn;
char maze[1005][1005];
bool mark[1005][1005];
bool flag;
int dx[5] {0 , 0 , 1 , 0 , -1};
int dy[5] {0 , 1 , 0 , -1 , 0};

bool check (int x , int y)
{
    if (x >= 1 && x <= n && y >= 1 && y <= m&& maze[x][y] == '.' && mark[x][y] == false)
    {
        return true;
    }
    return false;
}

void escape (int x , int y)
{
    if (x == n && y == m)
    {
        flag = true; 
        return ;
    }
    for (int i = 1; i <= 4; i++)
    {
        if (check(x + dx[i] , y + dy[i]))
        {
            mark[x + dx[i]][y + dy[i]] = true;
            escape(x + dx[i] , y + dy[i]);
        }
    }
}

int main()
{
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= m; j++)
        {
            cin >> maze[i][j];
        }
    }
    mark[1][1] = true;
    if (maze[1][1] == '#' || maze[n][m] == '#')
    {
        cout<<"No";
        return 0;
    }
    escape(1 , 1);
    if (flag == true) cout<<"Yes";
    else cout<<"No";
    return 0;
 }