空格样例HDU1241：Oil Deposits

每日一贴,今天的内容关键字为空格样例

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.

每日一道理
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Sample Input

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

Sample Output

原来第四个样例的第二个五后还有个空格，难怪输出不对劲，去掉空格就能够了，或者把%*C改成\n

#include <stdio.h>
#include <string.h>
char map[105][105],mat[105][105];
int vis[105][105];

void dfs(int i,int j)
{
    if(vis[i][j] || mat[i][j] == '*')
    return ;
    vis[i][j] = 1;
    dfs(i-1,j-1);
    dfs(i-1,j);
    dfs(i-1,j+1);
    dfs(i,j-1);
    dfs(i,j+1);
    dfs(i+1,j-1);
    dfs(i+1,j);
    dfs(i+1,j+1);
}

int main()
{
    int n,m;
    while(~scanf("%d%d%*c",&n,&m))
    {
        memset(map,0,sizeof(map));
        memset(mat,'*',sizeof(mat));
        memset(vis,0,sizeof(vis));
        int i,j,cnt = 0;
        if(!m&&!n)
        break;
        for(i = 0;i<n;i++)
        {
            for(j = 0;j<m;j++)
            {
                scanf("%c",&map[i][j]);
                mat[i+1][j+1] = map[i][j];
            }
            getchar();
        }
        for(i = 1 ;i<=n;i++)
        {
            for(j = 1;j<=m;j++)
            {
                if(!vis[i][j] && mat[i][j] == '@')
                {
                    dfs(i,j);
                    cnt++;
                }
            }
        }
        printf("%d\n",cnt);
    }
    return 0;
}

文章结束给大家分享下程序员的一些笑话语录： Borland说我很有前途，Sun笑了；Sun说我很有钱，IBM笑了；IBM说我很专业，Sybase笑了；Sybase说我数据库很牛，Oracle笑了；Oracle说我是开放的，Linux笑了；Linux说我要打败Unix，微软笑了；微软说我的系统很稳定，我们都笑了。

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空格和样例
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