P

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 

 

题意：
@代表细胞，如果两个@相邻或者对角则为同一种细胞，找到图中有多少种

分析：

深搜就行

代码：

#include<iostream>#include<cstring>#include<string>using namespace std;int a[105][105],n,m,b[16]={-1,-1,-1,0,-1,1,0,-1,0,1,1,-1,1,0,1,1};void dfs(int x,int y){    int i;    //cout<<"ssssyyy"<<endl;    for(i=0;i<16;i+=2)    {        if(x+b[i]>=0&&x+b[i]<m&&y+b[i+1]>=0&&y+b[i+1]<n&&a[x+b[i]][y+b[i+1]]==0)        {            //cout<<"ssss"<<endl;            a[x+b[i]][y+b[i+1]]=1;            dfs(x+b[i],y+b[i+1]);        }    }    return;}int main(){    int i,j,ans,k,l;    char f;    while(cin>>m>>n&&n&&m)    {        memset(a,0,sizeof(a));        for(i=0;i<m;i++)        for(j=0;j<n;j++)        {            cin>>f;            if(f=='*')            a[i][j]=1;        }        for(i=0,ans=0;i<m;i++)        for(j=0;j<n;j++)        {            if(a[i][j]==0)            {                dfs(i,j);                a[i][j]=1;                ans++;            }        }        cout<<ans<<endl;    }}

感受：

这个题我做了很长时间，第一次把移动的方向移到一个一维数组中，把控制的数目16 写成了8，找错找了好长时间