Oil Deposits

原题链接

题解

题目是说有多少个连通块，每一次以一个开始，遍历并标记他能够到达的点（就是同一个连通块中的点）

代码如下

dfs

#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <string>
#include <vector>
#include <stack>
#include <map>
#include <queue>
#include <sstream>
#include <set>
#include <cctype>
#define mem(a,b) memset(a,b,sizeof(a))

using namespace std;

typedef pair<int, int> PII;
const int N = 101;
int m, n;
char g[N][N];
int dx[8] = {0,0,1,1,1,-1,-1,-1}, dy[8] = {1,-1,0,1,-1,0,1,-1};

void dfs(int a, int b){
    for(int i = 0; i < 8; ++i){
        int x = a + dx[i], y = b + dy[i];
        if(x >= 0 && x < m && y >= 0 && y < n && g[x][y] == '@'){
            g[x][y] = '*';
            dfs(x, y);
        }
    }
}

int main(){

    while(cin >> m >> n){
        if(m == 0) break;
        for(int i = 0; i < m; ++i){
            for(int j = 0; j < n; ++j)
            cin >> g[i][j];
        }

        int res = 0;

        for(int i = 0; i < m; ++i){
            for(int j = 0; j < n; ++j){
                if(g[i][j] == '*') continue;
                res++;
                dfs(i, j);
            }
        }
        cout << res << endl;
    } 

    return 0;
}

bfs

#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <string>
#include <vector>
#include <stack>
#include <map>
#include <queue>
#include <sstream>
#include <set>
#include <cctype>
#define mem(a,b) memset(a,b,sizeof(a))

using namespace std;

typedef pair<int, int> PII;
const int N = 101;
char g[N][N];
int dx[8] = {0,0,1,1,1,-1,-1,-1}, dy[8] = {1,-1,0,1,-1,0,1,-1};

int main(){
    int m, n;
    while(cin >> m >> n){
        if(m == 0) break;
        for(int i = 0; i < m; ++i){
            for(int j = 0; j < n; ++j)
            cin >> g[i][j];
        }

        int res = 0;

        for(int i = 0; i < m; ++i){
            for(int j = 0; j < n; ++j){
                if(g[i][j] == '*') continue;
                queue<PII> q; res ++;
                q.push({i, j}); g[i][j] = '*';
                
                while(q.size()){
                    auto t = q.front(); q.pop();
                    for(int k = 0; k < 8; ++k){
                        int x = t.first + dx[k], y = t.second + dy[k];
                        if(x >= 0 && x < m && y >= 0 && y < n && g[x][y] == '@'){
                            q.push({x, y}); g[x][y] = '*';
                        }
                    }
                }

            }
        }
        cout << res << endl;
    } 

    return 0;
}

这个题目还可以用并查集做，以后水平提高了补上