第四个点超时，用dfs

ansan @ 2024-01-22 15:41:12

using namespace std;
const int N = 50;
bool vis[N][N];
int dx[4] = {1, -1, 0, 0};
long long dy[4] = {0, 0, 1, -1};
long long cha[50][50];
long long a[50][50];
long long n;
void dfs(int x, int y){
    for(int i = 0; i < 4; i++){
        int xx = x + dx[i], yy = y + dy[i];
            if(xx >= 0 && xx <= n+1 && yy >= 0 && yy <= n+1 && !vis[xx][yy]){
                vis[xx][yy] = 1;
                cha[xx][yy] = -1;
                dfs(xx,yy);
                vis[xx][yy] = 0;
            }
    }
    return;
}
int main(){
    cin >> n;
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= n; j++){
            cin >> a[i][j];
            if(a[i][j] == 1)
            vis[i][j] = 1;
        } 
    }
    vis[0][0] = 1;
    dfs(0,0);
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= n; j++){
            if(a[i][j] == 0 && cha[i][j] != -1)
            a[i][j] = 2;
            cout << a[i][j] << " ";
        }
        cout << "\n";
    }
    return 0;
}

by mooktian @ 2024-01-23 10:34:08

我帮你改了下，你自己看看。

#include <bits/stdc++.h>
#define ll long long
using namespace std;
const int N = 50;
bool vis[N][N];
int dx[4] = {1, -1, 0, 0};//方向下上左右 
int dy[4] = {0, 0, 1, -1};
//long long cha[50][50];
int a[50][50];
int n;
void dfs(int x, int y){
    for(int i = 0; i < 4; i++){
        int xx = x + dx[i], yy = y + dy[i];
            if(xx >= 0 && xx <= n+1 && yy >= 0 && yy <= n+1 && !vis[xx][yy] && a[xx][yy] != 1){//a[xx][yy] =1可以看成是墙，就不走 
                vis[xx][yy] = 1;
                //cha[xx][yy] = -1;
                dfs(xx,yy);
                //给矩阵外围多加一圈0，对于每个为0的点，只走一次，所以不需要回溯 
                //vis[xx][yy] = 0;
            }
    }
    return;
}
int main(){
    cin >> n;
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= n; j++){
            cin >> a[i][j];
            //if(a[i][j] == 1)
            //vis[i][j] = 1;
        } 
    }
    vis[0][0] = 1;
    dfs(0,0);
    for(int i = 1; i <= n; i++){
        for(int j = 1; j <= n; j++){
            //if(a[i][j] == 0 && cha[i][j] != -1)
            if(a[i][j] == 0 && vis[i][j] == 0)
            a[i][j] = 2;
            cout << a[i][j] << " ";
        }
        cout << "\n";
    }
    return 0;
}