bfs 练习题,简单bfs
题意:给一块地图,找出油田的块的数量,这里要考虑油田的八个方向,上下左右(左右)上(左右)下,存在则可以并在一起。@是油田,*是土地,m是行,n是列。
解题思路:用一个二维数组表示8个方向,然后bfs即可。
#include <iostream> #include <cstdio> #include <queue> #include <cstring> using namespace std; const int M = 105; char map[M][M];int m,n; int res; struct Oil { int x,y; }; queue <Oil> oil; int dire[8][2] = {{-1,0},{0,1},{1,0},{0,-1},{1,1},{-1,-1},{1,-1},{-1,1}}; //方向 void bfs() { while (!oil.empty()) { Oil oo = oil.front(); oil.pop(); int dx = oo.x; int dy = oo.y; for (int i = 0; i < 8; i++) { int x = dx+dire[i][0]; int y = dy+dire[i][1]; if (x >= 0 && x < m && y < n && y >= 0 && map[x][y] == '@') { map[x][y] = '*'; Oil o; o.x = x; o.y = y; oil.push(o); } } } } int main() { while (cin >> m >> n && m) { res = 0; memset(visited,0,sizeof (visited)); for (int i = 0;i < m;i++) for (int j = 0;j < n;j++) cin >> map[i][j]; for (int j = 0; j < m; j++) { for (int k = 0; k < n; k++) { if (map[j][k] == '@') { map[j][k] = '*'; Oil oo; oo.x = j; oo.y = k; oil.push(oo); res++; bfs(); } } } cout << res << endl; } return 0; }