[HDU1241] C

Oil Deposits

 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

此题为求油田的数量，连在一起的@为一个，只要在四周都可以

bfs代码

#include<cstdio>#include<cstring>#include<cmath>#include<queue>#include<algorithm>using namespace std;int m,n,ans;      //开头声明变量const int MAX = 1e2+10;typedef long long LL;char s[MAX][MAX];int vis[MAX][MAX];int fx[8]={0,0,1,-1,1,-1,1,-1};int fy[8]={1,-1,0,0,1,1,-1,-1};struct oil{    int x,y;};void bfs(int x,int y){    vis[x][y]=1;    queue<oil> q;       //队列    oil p;    p.x=x;p.y=y;    q.push(p);    while(!q.empty())    {        p=q.front();        q.pop();        for(int i=0;i<8;i++)   //八个方向        {            int xx=p.x+fx[i],yy=p.y+fy[i];            for(int j=0;j<n;j++)            {                if(xx>=0&&xx<m&&yy>=0&&yy<n&&!vis[xx][yy]&&s[xx][yy]=='@')                {                    oil w;                    vis[xx][yy]=1;                    w.x=xx;w.y=yy;                    q.push(w);                }            }        }    }}int main(){    while(~scanf("%d%d",&m,&n),m)    {        memset(vis,0,sizeof(vis));    //初始化为0        ans=0;        for(int i=0;i<m;i++)        {            scanf("%s",&s[i]);        }        for(int i=0;i<m;i++)        {            for(int j=0;j<n;j++)            {                if(!vis[i][j]&&s[i][j]=='@')                {                ans++;bfs(i,j);                }            }        }        printf("%d\n",ans);    }return 0;}

栈和队列知识

dfs

#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespace std;int m,n,ans;const int MAX = 1e2+10;typedef long long LL;char s[MAX][MAX];int vis[MAX][MAX];int fx[8]={0,0,1,-1,1,-1,1,-1};int fy[8]={1,-1,0,0,1,1,-1,-1};void dfs(int x,int y ){    vis[x][y]=1;    for(int i=0;i<8;i++)    {        int xx=x+fx[i],yy=y+fy[i];        if(xx>=0&&xx<m&&yy>=0&&yy<n&&!vis[xx][yy]&&s[xx][yy]=='@')        {            dfs(xx,yy);        }    }}int main(){    while(~scanf("%d%d",&m,&n),m)    {        memset(vis,0,sizeof(vis));        ans=0;        for(int i=0;i<m;i++)        {            scanf("%s",&s[i]);        }        for(int i=0;i<m;i++)        {            for(int j=0;j<n;j++)            {                if(!vis[i][j]&&s[i][j]=='@')                {                ans++;dfs(i,j);                }            }        }        printf("%d\n",ans);    }return 0;}