鏈接 : Here!

思路 : 搜索判斷連通塊個數, 所以 $DFS$ 或則 $BFS$ 都行嘍...., 首先記錄一下整個地圖中所有$Oil$的個數, 然後遍歷整個地圖, 從油田開始搜索它所能連通多少塊其他油田, 只需要把它所連通的油田個數減去, 就ok了

/*************************************************************************
    > File Name: E.cpp
    > Author: 
    > Mail: 
    > Created Time: 2017年11月26日 星期日 10時51分05秒
 

 ************************************************************************/

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;

#define MAX_N 150
int n, m;
int total_num;
int  vis[MAX_N][MAX_N];
int dx[8] = {0, 0, -1, 1, 1, -1, 1, -1};
int dy[8] = {-1, 1, 0, 0
 
, 1, -1, -1, 1};
char G[MAX_N][MAX_N];

void dfs(int x, int y) {
    vis[x][y] = 1;
    for (int i = 0 ; i < 8 ; ++i) {
        int now_x = x + dx[i];
        int now_y = y + dy[i];
        if (vis[now_x][now_y]) continue;
        if (now_x < 0 || now_x >= n || now_y < 0 || now_y >= m) continue
 
;
        if (G[now_x][now_y] != '@') continue;
        vis[now_x][now_y] = 1;
        --total_num;
        dfs(now_x, now_y);
    }
}
void solve() {
    for (int i = 0 ; i < n ; ++i) {
        for (int j = 0 ; j < m ; ++j) {
            if (G[i][j] == '@') {
                ++total_num;
            }
        }
    }
    memset(vis, 0, sizeof(vis));
    for (int i = 0 ; i < n ; ++i) {
        for (int j = 0 ; j < m ; ++j) {
            if (G[i][j] != '@' || vis[i][j]) continue;
            dfs(i, j);
        }
    }
}
int main() {
    while (scanf("%d%d", &n, &m) != EOF) {
        if (n == 0 && m == 0) break;
        memset(G, 0, sizeof(G));
        total_num = 0;
        for (int i = 0 ; i < n ; ++i) {
            getchar();
            scanf("%s", G[i]);    
        }
        solve();
        printf("%d\n", total_num);
    }
    return 0;
}

HDU 1241 Oil Deposits (DFS or BFS)