L - Oil Deposits HDU - 1241（BFS求连通分量）

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博客围绕HDU - 1241题目展开，该题要求在给定网格中计算不同油藏数量，即联通分量个数。输入为网格信息，输出是联通分量数量。解题思路是遍历每块田地，对未标记的油田进行BFS并标记联通分量，每次BFS联通分量加一。

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L - Oil Deposits HDU - 1241

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.
Input
The input file 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.
Output
For each grid, output the number of distinct oil deposits. Two different pockets are part of the same oil deposit if they are adjacent horizontally, vertically, or diagonally. An oil deposit will not contain more than
100 pockets.

Sample Input

1 1
*
3 5
@@*
@
@@*
1 8
@@***@
5 5
****@
@@@
@**@
@@@@
@@**@
0 0

Sample Output

0
1
2
2

题意：在一块田里，找到有几个联通分量，一块油地的上下左右左上…8个方向相邻的都算一个联通分量。

思路：便利每一块田地，当这一块是油田并且没有标记，就对这一块油田bfs，并标记他的联通分量，每次进行bfs联通分量就加一。

代码：

#include <iostream>
#include <cstring>
#include <queue>

using namespace std;

typedef struct node {
    int x,y;
} ty;

char a[102][102];
int via[102][102];
int n,m;
int ans;
ty t,d;

void bfs( int x, int y )
{
    int next[8][2] = {
        {1,0},{-1,0},{0,1},{0,-1},{1,1},{-1,-1},{1,-1},{-1,1}
    };
    ans ++;
    queue <ty> Q;
    t.x = x;
    t.y = y;
    Q.push(t);
    via[t.x][t.y] = 0;

    while ( !Q.empty() ) {
        t = Q.front();
        Q.pop();

        for ( int i=0; i<8; i++ ) {
            d.x = t.x + next[i][0];
            d.y = t.y + next[i][1];

            if ( d.x<0 || d.y<0 || d.x>=n || d.y>=m ) {
                continue ; // 事实证明，这条千万别漏写了，会wa
            }
            
            if ( a[d.x][d.y]=='@' && via[d.x][d.y]==0 ) {
                via[d.x][d.y] = 1;
                Q.push(d);
            }
        }
    }
}

int main()
{
    int i,j;
    while ( cin>>n>>m && n!=0 && m!=0 ) {
        for ( i=0; i<n; i++ ) {
            for ( j=0; j<m; j++ ) {
                cin >> a[i][j];
            }
        }

        ans = 0;
        memset(via,0,sizeof(via));
        for ( i=0; i<n; i++ ) { // 便利所有田
            for ( j=0; j<m; j++ ) {
                if ( a[i][j]=='@' && via[i][j]==0 ) {
                    bfs(i,j);
                }
            }
        }
        cout << ans << endl;

    }

    return 0;
}