如下所示的Map中,0代表海水,1代表岛屿,其中每一个岛屿与其八领域的区间的小岛能相连组成岛屿群。写代码,统计Map中岛屿个数。
/* Q1. Map [ 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 ] */
先直接上代码,后续等我有时间再写解题报告。
#include<iostream>#include<queue>using namespace std;typedef struct { int i; int j;
}position;void search(int a[][], int n, int i, int j, int cnt) { queue<position> qu = new queue<position>();
position p;
p.i = i;
p.j = j;
qu.push(p);
a[i][j] = cnt; while (!qu.empty()) {
p = qu.pop(); for (int ii = p.i - 1; ii <= p.i + 1; ii++) { for (int jj = p.j - 1; jj <= p.j + 1; jj++) { if (ii >= 0 && ii < n && jj >= 0 && jj < n && a[ii][jj] == 1 && (ii != i || jj != j)) {
a[ii][jj] = cnt;
p.i = ii;
p.j = jj;
qu.push(p);
}
}
}
}
}int count(int a[][], int n) { int cnt = 1; for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { if (a[i][j] == 1) {
cnt++; // 发现一个新陆地
search(a, n, i, j, cnt);
}
}
} return cnt;
}int main() { int n; cin >> n; int a[][] = new int[n][n]; for (int i = 0; i < n; i++) { for (int j = 0; j < n; j++) { cin >> a[i][j];
}
} int cnt = count(a, n); cout << cnt - 1 << endl; return 0;
}