方法一 dfs
#include <iostream>
#include <cstring>
#include <vector>
#include <queue>
using namespace::std;
typedef pair<int, int> pii;
const int dir[8][2] = { {-1, -1}, {-1, 0}, {-1, 1},
{0, -1}, {0, 1},
{1, -1}, {1, 0}, {1, 1} };
// 簡單的老題,BFS
int m, n;
char grid[101][101];
int bfs() {
bool visited[101][101] = { 0 };
int count = 0;
queue<pii> qu;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (grid[i][j] == '@' && visited[i][j] == 0) {
visited[i][j] = 1;
qu.push(pii(i, j));
while (!qu.empty()) {
int x = qu.front().first, y = qu.front().second;
for (int op = 0; op < 8; ++op) {
int n_x = x + dir[op][0], n_y = y + dir[op][1];
if (n_x >= 0 && n_x < m && n_y >= 0 && n_y < n
&& visited[n_x][n_y] == 0 && grid[n_x][n_y] == '@') {
visited[n_x][n_y] = 1;
qu.push(pii(n_x, n_y));
}
}
qu.pop();
}
count++;
}
}
}
return count;
}
int main(void) {
while ((cin >> m >> n) && m != 0 && n!= 0) {
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
cin >> grid[i][j];
}
}
cout << bfs() << endl;
}
return 0;
}
方法二 dfs
#include <iostream>
#include <cstring>
#include <vector>
#include <queue>
using namespace::std;
typedef pair<int, int> pii;
const int dir[8][2] = { {-1, -1}, {-1, 0}, {-1, 1},
{0, -1}, {0, 1},
{1, -1}, {1, 0}, {1, 1} };
// 再試試DFS的方法
int m, n;
char grid[101][101];
bool visited[101][101];
// 找到和(x,y)相連的所有方塊
bool dfs(int x, int y) {
if (grid[x][y] == '*' || visited[x][y] == 1) return false;
visited[x][y] = 1;
for (int op = 0; op < 8; ++op) {
int n_x = x + dir[op][0], n_y = y + dir[op][1];
if (n_x >= 0 && n_x < m && n_y >= 0 && n_y < n
&& visited[n_x][n_y] == 0 && grid[n_x][n_y] == '@') {
dfs(n_x, n_y);
}
}
return true;
}
int main(void) {
while ((cin >> m >> n) && m != 0 && n!= 0) {
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
cin >> grid[i][j];
}
}
memset(visited, 0, sizeof(bool) * 101 * 101);
int ans = 0;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (dfs(i, j)) ans++;
}
}
cout << ans << endl;
}
return 0;
}