poj2386（简单dfs）

就是求图中有多少个水洼。对图进行dfs遍历，并把是水洼的地方全部标记。然后从下一个是水哇的地方再进行dfs。

#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
int graph[105][105];
int sum;
bool vis[105][105];
int dx[] = {-1, -1, -1,  0, 0,  1, 1, 1};
int dy[] = {-1,  0,  1, -1, 1, -1, 0 ,1};
int row, col;

bool check(int x, int y)
{
    if( x >= 0 && x < row && y >= 0 && y < col)
        return true;
    return false;
}

void dfs(int x, int y)
{
    int ix, iy;
    for(int i = 0; i < 8; i++)
    {
        ix = x + dx[i];
        iy = y + dy[i];
        if(!vis[ix][iy] && check(ix, iy) && graph[ix][iy] == 1)
        {
            vis[ix][iy] = true;
            dfs(ix, iy);
        }
    }
}
int main()
{
    //freopen("in.txt", "r", stdin);
    cin >> row >> col;
    getchar();
    memset(vis, false, sizeof(vis));

    char c;
    for(int i = 0; i < row; i ++)
    {
        for(int j = 0; j < col; j++)
        {
            c = getchar();
            graph[i][j] = (c == '.') ? 0 : 1;
        }
        getchar();
    }
    sum = 0;

    for(int i  = 0; i < row; i ++)
    {
        for(int j = 0; j < col; j++)
        {
            if(graph[i][j] == 1 && vis[i][j] == false)
            {
                sum++;
                vis[i][j] = true;
                dfs(i, j);
            }
        }
    }

    cout << sum << endl;
    return 0;
}