代码随想录 99. 岛屿数量

99. 岛屿数量

#include<bits/stdc++.h>
using namespace std;
 
void dfs(vector<vector<int>>& mp, vector<vector<int>>& visit, int y, int x){
    if (visit[y][x] == 1 || mp[y][x] == 0) return;
    visit[y][x] = 1;
    if (y > 0) dfs(mp, visit, y - 1, x);
    if (x > 0) dfs(mp, visit, y, x - 1);
    if (y < mp.size() - 1) dfs(mp, visit, y + 1, x);
    if (x < mp[0].size() - 1) dfs(mp, visit, y, x + 1);
}
 
int main(){
    int n, m;
    cin >> n >> m;
    vector<vector<int>> mp(n, vector<int>(m, 0));
    vector<vector<int>> visit(n, vector<int>(m,0));
    int result = 0;
     
    for (int i = 0; i < n; i++){
        for (int j = 0; j < m; j++){
            cin >> mp[i][j];
        }
    }
    for (int i = 0; i < n; i++){
        for (int j = 0; j < m; j++){
            if (mp[i][j] == 1 && visit[i][j] == 0){
                result++;
                dfs(mp, visit, i, j);
            }
        }
    }
    cout << result;
    return 0;
}

这是dfs的版本，其实就是在遇到一个没访问过的1时，就将所有能和这个1组成一个岛的1全部设为异界访问过，dfs与bfs只是采取两种遍历岛内元素的方法

#include<bits/stdc++.h>
using namespace std;
 
void bfs(vector<vector<int>>& mp, vector<vector<int>>& visit, int y, int x){
    queue<pair<int, int>> q;
    q.push({y, x});
    visit[y][x] = 1;
    while(!q.empty()){
        pair<int, int> tmp = q.front();
        int tmpy = tmp.first, tmpx = tmp.second;
        q.pop();
        if (tmpy > 0){
            if (visit[tmpy - 1][tmpx] == 0 && mp[tmpy - 1][tmpx] == 1){
                q.push({tmpy - 1, tmpx});
                visit[tmpy - 1][tmpx] = 1;
            }
        }
        if (tmpx > 0){
            if (visit[tmpy][tmpx - 1] == 0 && mp[tmpy][tmpx - 1] == 1){
                q.push({tmpy, tmpx - 1});
                visit[tmpy][tmpx - 1] = 1;
            }
        }
        if (tmpy < mp.size() - 1){
            if (visit[tmpy + 1][tmpx] == 0 && mp[tmpy + 1][tmpx] == 1){
                q.push({tmpy + 1, tmpx});
                visit[tmpy + 1][tmpx] = 1;
            }
        }
        if (tmpx < mp[0].size() - 1){
            if (visit[tmpy][tmpx + 1] == 0 && mp[tmpy][tmpx + 1] == 1){
                q.push({tmpy, tmpx + 1});
                visit[tmpy][tmpx + 1] = 1;
            }
        }
    }
}
 
int main(){
    int n, m;
    cin >> n >> m;
    vector<vector<int>> mp(n, vector<int>(m, 0));
    vector<vector<int>> visit(n, vector<int>(m,0));
    int result = 0;
     
    for (int i = 0; i < n; i++){
        for (int j = 0; j < m; j++){
            cin >> mp[i][j];
        }
    }
    for (int i = 0; i < n; i++){
        for (int j = 0; j < m; j++){
            if (mp[i][j] == 1 && visit[i][j] == 0){
                result++;
                bfs(mp, visit, i, j);
            }
        }
    }
    cout << result;
    return 0;
}

这是bfs的版本，需要强调的是，标注已访问需要在入队时就标注，出队时标注仍然会有重复访问导致超时的可能。

代码随想录 99. 岛屿数量