• 박인서

# 3. 코드 ¶

## 3.1. 15이원준 ¶

```#include<iostream>
#include<vector>
#include<algorithm>

using namespace std;

int arr[27][27] = { 0, };
int N, num = 0;
vector<int> v;

void search(int i, int j) {
v[num]++;
arr[i][j] = -1;
if(arr[i+1][j] == 1)
search(i + 1, j);
if(arr[i-1][j] == 1)
search(i - 1, j);
if(arr[i][j+1] == 1)
search(i, j + 1);
if(arr[i][j-1] == 1)
search(i, j - 1);
}

int main() {
cin >> N;
for (int i = 0; i<27; i++) {
for (int j = 0; j<27; j++) {
arr[i][j] = 0;
}
}

for (int i = 1; i <= N; i++) {
for (int j = 1; j <= N; j++) {
int tmp;
scanf("%1d", &tmp);
arr[i][j] = tmp;
}
}
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= N; j++) {
if (arr[i][j] == 1) {
v.push_back(0);
search(i, j);
num++;
}
}
}
sort(v.begin(), v.end());
cout<<v.size()<<endl;
for (int i = 0; i<v.size(); i++) {
cout << v[i] << endl;
}
}
```

## 3.2. 박인서 ¶

```#include <iostream>
#include <vector>
#include <algorithm>
char a[30][30];
bool visit[30][30];
int n;
std::vector<int> res;

int dfs (int i,int j){
int cnt=1;
visit[i][j]=true;

if (i!=1 && a[i-1][j]=='1' && !visit[i-1][j])
cnt+=dfs(i-1,j);
if (i!=n && a[i+1][j]=='1' && !visit[i+1][j])
cnt+=dfs(i+1,j);
if (j!=1 && a[i][j-1]=='1' && !visit[i][j-1])
cnt+=dfs(i,j-1);
if (j!=n && a[i][j+1]=='1' && !visit[i][j+1])
cnt+=dfs(i,j+1);

return cnt;
}

int main (){
std::cin>>n;
for (int i=1;i<=n;i++){
for (int j=1;j<=n;j++){
std::cin>>a[i][j];
}
}

for (int i=1;i<=n;i++){
for (int j=1;j<=n;j++){
if (a[i][j]=='1' && !visit[i][j])
res.push_back(dfs(i,j));
}
}

std::cout<<res.size()<<std::endl;
std::sort(res.begin(),res.end());
for(int i=0;i<res.size();i++)
std::cout<<res[i]<<std::endl;

return 0;
}
```

# 4. 아이디어 ¶

## 4.2. 박인서 ¶

• 단지 중 1곳 파악->단지 완전 탐색 순서로 단지를 탐색하면 된다