hdu 1241 Oil Deposits（搜索基础题）

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1241

Oil Deposits

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 25435    Accepted Submission(s): 14636

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

0122

 

Source

Mid-Central USA 1997

 

Recommend

Eddy

 

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题目大意：求总共油田的个数，@代表油田，*代表这块地是空的

解析：简单DFS，就是让它递归找下上下左右，还有斜上方斜下方，注意边界问题

代码如下：

#include<iostream>#include<algorithm>#include<map>#include<stack>#include<queue>#include<vector>#include<set>#include<string>#include<cstdio>#include<cstring>#include<cctype>#include<cmath>#define N 4009using namespace std;const int inf = 1e9;const int mod = 1<<30;const double eps = 1e-8;const double pi = acos(-1.0);typedef long long LL;int mp[110][110], m, n;int nx[11] = {-1, 0, 0, 1, -1, -1, 1, 1};int ny[11] = {0, -1, 1, 0, -1, 1, -1, 1};void dfs(int i, int j){    int k, x, y;    for(k = 0; k < 8; k++)    {        x = i + nx[k]; y = j + ny[k];        if(x < 1 || y < 1 || x > m || y > n) continue;        if(!mp[x][y]) continue;        mp[x][y] = 0;        dfs(x, y);    }}int main(){    int i, j;    char ch;    while(scanf("%d%d", &m, &n), n + m)    {        memset(mp, 0, sizeof(mp));        int ans = 0;        for(i = 1; i <= m; i++)        {            for(j = 1; j <= n; j++)            {                scanf(" %c", &ch);                if(ch == '@') mp[i][j] = 1;            }        }        for(i = 1; i <= m; i++)        {            for(j = 1; j <= n; j++)            {                if(mp[i][j]) ans++, dfs(i, j);            }        }        printf("%d\n", ans);    }    return 0;}