101. 孤岛的总面积

[Jay-Chou118]

这道题目的意思应该是计算孤岛个数吧，根据题意，
对于
5 6
1 1 0 0 0 0
1 1 0 0 0 0
0 0 1 1 0 0
0 0 0 0 0 1
这个输入，应该输出是1.
但是按照卡哥的代码，输出会是2，问题就出在应该是找到==1的陆地后，继续进行dfs，而不是直接count++

[Jay-Chou118]

`#include
#include
using namespace std;
int dir[4][2] = {{-1, 0}, {0, -1}, {1, 0}, {0, 1}};
void dfs(vector<vector>& grid, int x, int y) {
grid[x][y] = 0;
for (int i = 0; i < 4; i++) { // 向四个方向遍历
int nextx = x + dir[i][0];
int nexty = y + dir[i][1];
// 超过边界
if (nextx < 0 || nextx >= grid.size() || nexty < 0 || nexty >= grid[0].size()) continue;
// 不符合条件，不继续遍历
if (grid[nextx][nexty] == 0) continue;

    dfs(grid, nextx, nexty);
}
return;

}
int main() {
int n, m;
cin >> n >> m;
vector<vector> grid(n, vector(m, 0));
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
cin >> grid[i][j];
}
}
// 从左侧边和右侧边向中间遍历
for (int i = 0; i < n; i++) {
if (grid[i][0] == 1) dfs(grid, i, 0);
if (grid[i][m - 1] == 1) dfs(grid, i, m - 1);
}
// 从上边和下边向中间遍历
for (int j = 0; j < m; j++) {
if (grid[0][j] == 1) dfs(grid, 0, j);
if (grid[n - 1][j] == 1) dfs(grid, n - 1, j);
}
int count = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++) {
if (grid[i][j] == 1) {
count++;
dfs(grid,i,j);
}
}
}
cout << count << endl;
}`