深搜之入门例题 DFS Oil Deposits

Oil Deposits

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 21383 Accepted Submission(s): 12319


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

题意：@是一个油田，如果它的周围8个方向有一个是油田的话，那么这两个油田是一块。问：这样的油田有多少块？

思路：因为这里的每一个地方都需要搜索判断所以采用深搜。

#include <stdio.h>#include <algorithm>#include <string.h>#include <math.h>#include <cmath>#include <queue>#include <iostream>#include <cmath>#define me(a) memset(a, 0,sizeof(a))using namespace std;int n, m;char ch[1000][1000];int move[8][2] = {1,0, -1,0, 0,1, 0,-1, 1,1, -1,-1, 1,-1, -1,1};void dfs(int x, int y){    int next_x, next_y;    ch[x][y] = '*';    for(int i = 0; i < 8; i++)    {        next_x = x + move[i][0];        next_y = y + move[i][1];        if(next_x >= 0 && next_x < n && next_y >= 0 && next_y < m)        {            if(ch[next_x][next_y] == '@')            {                dfs(next_x, next_y);            }        }    }}int main(){    int ans;    while(~scanf("%d%d", &n, &m)&&(n!=0||m!=0))    {        for(int i = 0; i < n; i++)            scanf("%s", ch[i]);        ans = 0;        for(int i = 0; i < n; i++)        {            for(int j = 0; j < m; j++)            {                if(ch[i][j] == '@')                {                    dfs(i, j);                    ans++;                }            }        }        printf("%d\n", ans);    }    return 0;}