探查油田——廣度搜索

/*
1562:Oil Deposits
  總時間限制: 1000ms 內存限制: 65536kB
描述
The GeoSurvComp geologic survey company is responsible for detecting underground oil deposits. GeoSurvComp works with one large rectangular region of land at a time, and creates a grid that divides the land into numerous square plots. It then analyzes each plot separately, using sensing equipment to determine whether or not the plot contains oil. A plot containing oil is called a pocket. If two pockets are adjacent, then they are part of the same oil deposit. Oil deposits can be quite large and may contain numerous pockets. Your job is to determine how many different oil deposits are contained in a grid.
輸入
The input contains one or more grids. Each grid begins with a line containing m and n, the number of rows and columns in the grid, separated by a single space. If m = 0 it signals the end of the input; otherwise 1 <= m <= 100 and 1 <= n <= 100. Following this are m lines of n characters each (not counting the end-of-line characters). Each character corresponds to one plot, and is either `*', representing the absence of oil, or `@', representing an oil pocket.

輸出
are adjacent horizontally, vertically, or diagonally. An oil deposit will not contain more than 100 pockets.
樣例輸入
1 1
*
3 5
*@*@*
**@**
*@*@*
1 8
@@****@*
5 5
****@
*@@*@
*@**@
@@@*@
@@**@
0 0
樣例輸出
0
1
2
2
來源
Mid-Central USA 1997
廣度搜索，設置節點隊列，8個方向搜索
清空隊列、初始化隊列
隊首元素探查後出隊
檢查入隊元素是否在地圖內
一個元素入隊時應作標記
 */
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define _N 100
struct Node
{
	int x,y;
	Node(){};
	Node(int x_,int y_):x(x_),y(y_){};
	bool isInMap(int m,int n)
	{
		return x < m && x >= 0 && y < n && y >= 0;
	}
}que[_N*_N];
int queHead,queTail;
char mat[_N][_N];
bool vis[_N][_N];
int m,n;
int dx[] = {-1, -1, -1,  0,  0,  1,  1,  1};
int dy[] = {-1,  0,  1, -1,  1, -1,  0,  1};
void bfs(int x, int y, int m, int n) {
    queHead = queTail = 0; // 清空隊列
    // 將(x, y)加入隊列
    que[queTail++] = Node(x, y);
    vis[x][y] = true;
    // 開始BFS搜索
    for (; queHead < queTail; ++queHead) {
        // 探查隊首元素周圍的8個位置
        for (int i = 0; i < 8; ++i) {
            // 根據偏移量計算位置
            Node nxt(que[queHead].x + dx[i], que[queHead].y + dy[i]);
            // 若該位置在地圖中且爲未訪問過的'@'，則加入隊列
            if (nxt.isInMap(m, n) && !vis[nxt.x][nxt.y] &&
                mat[nxt.x][nxt.y] == '@') {
                que[queTail++] = nxt;
                vis[nxt.x][nxt.y] = true;
            }
        }
    }
}

int main()
{

	while(cin >> m >> n)
	{
		int ret = 0;
		if(m == 0)
			break;
		for(int i = 0; i < m ;++i)
			for(int j = 0; j < n; ++j)
				cin >> mat[i][j];
		memset(vis, 0, sizeof(vis)); // 初始化標記
		// 遍歷整個地圖
		for (int i = 0; i < m; ++i) {
			for (int j = 0; j < n; ++j) {
				// 探查到了一個油礦
				if (mat[i][j] == '@' && !vis[i][j])
				{
					bfs(i, j, m, n);
					ret++;
				}
			}
		}
		cout << ret << endl;
	}
}

/*
遞歸版本，對於每個@的位置進行8個方向搜索，如果發現@將其標記
繼續搜索
 */
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
char mat[101][101];
bool vis[101][101];
int m,n;
int dx[] = {-1, -1, -1,  0,  0,  1,  1,  1};
int dy[] = {-1,  0,  1, -1,  1, -1,  0,  1};
void bfs(int x,int y)
{
	if(x < 0 || x >= m || y < 0 || y >= n)
		return;
	vis[x][y] = true;
	for(int i = 0; i < 8; ++i)
	{
		if(!vis[x+dx[i]][y+dy[i]] && mat[x+dx[i]][y+dy[i]] == '@')
			bfs(x+dx[i],y+dy[i]);
	}
}
int main()
{

	while(cin >> m >> n)
	{
		int ret = 0;
		if(m == 0)
			break;
		for(int i = 0; i < m ;++i)
			for(int j = 0; j < n; ++j)
				cin >> mat[i][j];
		memset(vis, 0, sizeof(vis)); // 初始化標記
		// 遍歷整個地圖
		for (int i = 0; i < m; ++i) {
			for (int j = 0; j < n; ++j) {
				// 探查到了一個油礦
				if (mat[i][j] == '@' && !vis[i][j])
				{
					bfs(i, j);
					ret++;
				}
			}
		}
		cout << ret << endl;
	}
}