求调

zengzeyu @ 2024-10-18 21:26:50

#include<bits/stdc++.h>
using namespace std;
char c[105][105];
int n,m;
bool dfs(int x,int y){
    if(c[x][y] == '#'){
        return false;
    }
    if(x == n && y == m){
        return true;
    }
    if(x < 1 || y < 1 || x > n || y > m){
        return false;
    }
    bool u = dfs(x-1,y),l = dfs(x,y-1),r = dfs(x,y+1),d = dfs(x+1,y);
    bool res = u || l || r || d;
    return res;
}
int main(){
    cin>>n>>m;
    for(int i = 1;i <= n;i ++){
        for(int j = 1;j <= m;j ++){
            cin>>c[i][j];
        }
    }
    bool tmpv = dfs(1,1);
    tmpv?cout<<"Yes":cout<<"No";
}

by kczw @ 2024-10-18 21:33:14

@zengzeyu 不难发现你的代码会在几个点来回走

by kczw @ 2024-10-18 21:35:07

@zengzeyu 你可以使用一个数组记录某个点是否被遍历过，如果已经被遍历过了，那么没有必要继续dfs，而是应该返回退出单次dfs并且返回假。

by zengzeyu @ 2024-10-18 21:37:20

@kczw 感谢大佬!

by kczw @ 2024-10-18 21:38:24

@zengzeyu AC 代码

#include<bits/stdc++.h>
using namespace std;
char c[105][105];
int n,m,vis[105][105];
bool dfs(int x,int y){
    if(c[x][y] == '#'||vis[x][y]==1//如果已经被遍历，没有必要二次遍历 
    ){
        return false;
    }
    if(x == n && y == m){
        return true;
    }
    if(x < 1 || y < 1 || x > n || y > m){
        return false;
    }
    vis[x][y]=1;//标记为已被遍历 
    bool u = dfs(x-1,y),l = dfs(x,y-1),r = dfs(x,y+1),d = dfs(x+1,y);
    bool res = u || l || r || d;
    return res;
}
int main(){
    cin>>n>>m;
    for(int i = 1;i <= n;i ++){
        for(int j = 1;j <= m;j ++){
            cin>>c[i][j];
        }
    }
    bool tmpv = dfs(1,1);
    tmpv?cout<<"Yes":cout<<"No";
}

by zengzeyu @ 2024-10-18 21:49:59

AC了,本贴结(顺便 %@kczw 大佬)

by zengzeyu @ 2024-10-28 17:56:02

已关