我的：
```cpp
#include<bits/stdc++.h>
using namespace std;
int n;
char maps[31][31] ;
int dir[4][2]={{0,1},{1,0},{-1,0},{0,-1}},vis[31][31];
struct node{
int x,y;
}s;
void bfs(int x,int y){
s.x = x;
s.y = y;
queue <node> que ;
que.push(s);
vis[x][y] = 1;
while(!que.empty()){
node now = que.front ();
que.pop();
for(int i=0;i<4;i++){
node next = now;
next.x+=dir[i][0];
next.y+=dir[i][1];
if(maps[next.x][next.y] == '1' || vis[next.x][next.y] == 1) continue;
else maps[next.x][next.y] = '2';
vis[next.x][next.y] = 1;
que.push(next);
}
}
}
int main(){
cin >> n;
memset (maps,'1',sizeof(maps));
for(int i=1;i<=n;i++){
for(int j=1;j<=n;j++){
cin>>maps[i][j];
}
}
for(int i=n;i>=1;i--){
for(int j=1;j<=n;j++){
if(maps[i][j] == '1') break;
else vis[i][j] = 1 ;
}
for(int j=n;j>=1;j--){
if(maps[i][j] == '1') break;
else vis[i][j] = 1 ;
}
}
for(int j=n;j>=1;j--){
for(int i=1;i<=n;i++){
if(maps[i][j] == '1') break;
else vis[i][j] = 1 ;
}
for(int i=n;i>=1;i--){
if(maps[i][j] == '1') break;
else vis[i][j] = 1 ;
}
}
int x=1,y=1,t=1;
for(int i=1;i<=n && t;i++){
for(int j = 1;j<=n && t;j++){
if(vis[i][j] == 0 && maps[i][j] == '0') {
x = i,y = j;
t = 0;
}
}
}
maps[x][y] = '2';
bfs(x,y);
for(int i = 1;i<=n;i++){
cout << maps[i][1];
for(int j = 2;j<=n;j++){
cout << " " <<maps[i][j];
}
cout << endl;
}
return 0;
}
```
by dingshengyang @ 2021-09-10 18:37:48
大水法
DFS + 骚操作
```cpp
#include <iostream>
#include <string>
using namespace std;
int a[600][600];
void dfs(int x,int y,int type){ //type代表将搜过的0标记的数字，搜外围就标2，防水空间就标1
a[x][y] = type;
if (a[x+1][y]!=0 && a[x-1][y]!=0 && a[x][y+1]!=0 && a[x][y-1]!=0){
return;
}
if (a[x + 1][y] == 0)dfs(x + 1,y,type);
if (a[x - 1][y] == 0)dfs(x - 1,y,type);
if (a[x][y + 1] == 0)dfs(x,y + 1,type);
if (a[x][y - 1] == 0)dfs(x,y - 1,type);
}
int main(){
int n;
cin>>n;
for (int i = 0;i <= n + 1;i++){ //越界保护，将所有边界标为2
for (int j = 0;j <= n + 1;j++){
a[i][j] = 2;
}
}
for (int i = 1;i <= n;i++){
for (int j = 1;j <= n;j++){
cin>>a[i][j];
}
}
// for (int i = 0;i <= n + 1;i++){
// for (int j = 0;j <= m + 1;j++) cout<<a[i][j];
// cout<<endl;
// }
int x = 2,y = 1;
int side = 1;
bool flag = false;
while (true){ //边界搜索，将边界搜到的空地标为2（即淹水状态）
if (flag == true && x == 1 && y == 1) break;
if (a[x][y] == 0){
dfs(x,y,2);
}
if (x == n && y == 1) side = 2;
if (x == n && y == n) side = 3;
if (x == 1 && y == n) side = 4;
if (side == 1) x++;
if (side == 2) y++;
if (side == 3) x--;
if (side == 4) y--;
flag = true;
}
for (int i = 0;i <= n;i++){
for (int j = 0;j <= n;j++){ //保护区搜索，空地标为1（即围墙）
if (a[i][j] == 0) dfs(i,j,3);
}
}
for (int i = 1;i <= n;i++){
for (int j = 1;j <= n;j++){
if (a[i][j] == 2) cout<<0<<" ";
else if (a[i][j] == 3) cout<<2<<" ";
else cout<<1<<" ";
}
cout<<endl;
}
```
by Escapism @ 2021-09-10 18:42:39
```cpp
#include<iostream>
#include<cstdio>
#include<queue>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxn=101;
int a[maxn][maxn];
int n;
int dx[4]={-1,0,1,0};
int dy[4]={0,1,0,-1};
struct node
{
int x;
int y;
};
queue <node> lq;
bool vis[maxn][maxn];
void bfs(int x,int y)
{
lq.push((node){x,y});
vis[0][0]=1;
while(!lq.empty())
{
for(int i=0;i<4;i++)
{
int xx=lq.front().x+dx[i];
int yy=lq.front().y+dy[i];
if(!vis[xx][yy]&&xx>=0&&xx<=n+1&&yy>=0&&yy<=n+1&&a[xx][yy]==0)
{
lq.push((node){xx,yy});
vis[xx][yy]=1;
}
}
lq.pop();
}
}
int main()
{
cin>>n;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
cin>>a[i][j];
}
}
bfs(0,0);
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
{
if(a[i][j]==0&&vis[i][j]==0)
{
cout<<2<<" ";
}
else cout<<a[i][j]<<" ";
}
cout<<endl;
}
return 0;
}
```
我的
by thatsgame @ 2021-09-10 18:46:21
用深搜过了
by Editzed @ 2021-09-10 18:48:35
话说不用结构体可能更好点。。。
by lamp_forever @ 2021-09-10 18:48:58
```cpp
#include<bits/stdc++.h>
using namespace std;
int xx[]={0,-1,0,1};
int yy[]={1,0,-1,0};
int mp[40][40];
bool vis[40][40];
int n,m;
queue<int>x;
queue<int>y;
int main()
{
cin>>n;
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
cin>>mp[i][j];
x.push(0);
y.push(0);
vis[0][0]=1;
while(!x.empty())
{
for(int i=0;i<4;i++)
{
int dx=x.front()+xx[i];
int dy=y.front()+yy[i];
if(dx>=0&&dx<=n+1&&dy>=0&&dy<=n+1&&mp[dx][dy]==0&&vis[dx][dy]==0)
{
x.push(dx);
y.push(dy);
vis[dx][dy]=1;
}
}
x.pop();
y.pop();
}
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
if(mp[i][j]==0&&vis[i][j]==0)
cout<<2<<" ";
else
cout<<mp[i][j]<<" ";
cout<<endl;
}
}
```
by lamp_forever @ 2021-09-10 18:49:35