DFS-HDU-1241-Oil Deposits

Oil Deposits

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18633 Accepted Submission(s): 10736

Problem Description
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

当初刚进软协的时候一个学长拿这个题考过我，当时只会C基础语法的我硬是憋了一个DFS出来，凑活着水吧。

//
// main.cpp
// 简单搜索-L-Oil Deposits
//
// Created by 袁子涵 on 15/8/17.
// Copyright (c) 2015年 袁子涵. All rights reserved.
//

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <stdlib.h>

using namespace std;

bool visit[101][101];
char land[101][101];
int sum;
int m,n;

void dfs(int x,int y)
{
    if (visit[x][y]) {
        return;
    }
    //先把自身消去
    land[x][y]='*';
    visit[x][y]=1;
    //上下左右斜对角
    if (x>0) {
        if (land[x-1][y]=='@') {
            dfs(x-1, y);
        }
        if (y>0 && land[x-1][y-1]=='@') {
            dfs(x-1, y-1);
        }
        if (y<n-1 && land[x-1][y+1]=='@') {
            dfs(x-1, y+1);
        }
    }
    if (x<m-1) {
        if (land[x+1][y]=='@') {
            dfs(x+1, y);
        }
        if (y>0 && land[x+1][y-1]=='@') {
            dfs(x+1, y-1);
        }
        if (y<n-1 && land[x+1][y+1]=='@') {
            dfs(x+1, y+1);
        }
    }
    if (y>0) {
        if (land[x][y-1]=='@') {
            dfs(x, y-1);
        }
    }
    if (y<n-1) {
        if (land[x][y+1]=='@') {
            dfs(x, y+1);
        }
    }
    return;
}

int main(int argc, const char * argv[]) {
    while (1) {
        cin >> m >> n;
        if (m==0) {
            return 0;
        }
        memset(land, 0, sizeof(land));
        memset(visit, 0, sizeof(visit));
        sum=0;
        for (int i=0; i<m; i++) {
            for (int j=0; j<n; j++) {
                cin >> land[i][j];
            }
        }
        for (int i=0; i<m; i++) {
            for (int j=0; j<n; j++) {
                if (visit[i][j]==0 && land[i][j]=='@') {
                    dfs(i, j);
                    sum++;
                }
            }
        }
        cout << sum << endl;
    }
    return 0;
}