不撞南墙不回头———深度优先搜索（DFS）Oil Deposits

Posted 2020-07-16 Timeashore

Oil Deposits

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 23018    Accepted Submission(s): 13272

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

 

 

 

 

刚刚学了深搜，做一下总结。

DFS

按着一条路走到，不撞南墙不回头，回的话也是回到上一次最近调用dfs()函数的地方，然后看看是否还有满足条件的，如果有则按满足条件的那个地方一直往下搜索（追根究底，一直往下，一直），直到撞南墙，深搜就是把所有的地方都遍历一次，一个也不放过

 

新手总结，不到指出请大家指正！！

 

附代码：

#include <stdio.h>
#include <string.h>
int m,n,total = 0,book[101][101];
char c[101][101];
int next[8][2]=
{
{0,1},//右
{1,1},//右下
{1,0},//下
{1,-1},//左下
{0,-1},//左
{-1,-1},//左上
{-1,0},//上
{-1,1}//右上
};

void dfs(int x,int y)
{
int tx,ty,k;
for(k=0;k<8;k++)
{
tx = x+next[k][0];
ty = y+next[k][1];

if(tx<1||ty<1||tx>m||ty>n)
{
continue;
}
if(c[tx][ty]==‘@‘&&book[tx][ty]==0)
{
//total++;//这里不能加1 ，因为是跟上一个连着的油田，都算是一块油田
book[tx][ty] = 1;
dfs(tx,ty);//8个中有一个满足条件，以这一个为中心，继续往下搜，如果不满足条件，返回来，接着上一次没搜完的搜
//book[tx][ty] = 0;//每一格就走一次，搜过了不再搜
}
}
return;
}
int main()
{
int i,j;
while(scanf("%d%d",&m,&n)&&m!=0)
{
getchar();
total = 0;
memset(book,0,sizeof(book));
for(i=1;i<=m;i++){
for(j=1;j<=n;j++){
scanf("%c",&c[i][j]);
}
getchar();
}
for(i=1;i<=m;i++)
{
for(j=1;j<=n;j++)
{
if(c[i][j]==‘@‘ && book[i][j]==0)
{
total++;//加它本身
dfs(i,j);//对它的八个方向进行搜索
}
}
}
printf("%d\n",total);
}
return 0;
}