深度优先搜索（DFS）

最新推荐文章于 2025-03-20 20:34:00 发布

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本文介绍了一种通过深度优先搜索算法来计算二维地图中湖泊数量的方法。针对给定的N*M矩阵，每个元素代表水('W')或陆地('.')，算法能够找出所有独立湖泊的数量。湖泊定义为相互连接的水域，采用八邻域连接方式。

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写在前面的话：搜索问题第一次接触，感觉很陌生，所以只有向大神请教！看看代码也不是很难，就是有一两处比较难懂，特此剪切，深度分析：不就快懂，只求能会。慢慢摸索吧！

A - Lake Counting

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

Submit Status

A - Lake Counting

Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u

Submit Status

Description

Due to recent rains, water has pooled in various places in Farmer John's field, which is represented by a rectangle of N x M (1 <= N <= 100; 1 <= M <= 100) squares. Each square contains either water ('W') or dry land ('.'). Farmer John would like to figure out how many ponds have formed in his field. A pond is a connected set of squares with water in them, where a square is considered adjacent to all eight of its neighbors.

Given a diagram of Farmer John's field, determine how many ponds he has.

Input

* Line 1: Two space-separated integers: N and M

* Lines 2..N+1: M characters per line representing one row of Farmer John's field. Each character is either 'W' or '.'. The characters do not have spaces between them.

Output

* Line 1: The number of ponds in Farmer John's field.

Sample Input

10 12
W........WW.
.WWW.....WWW
....WW...WW.
.........WW.
.........W..
..W......W..
.W.W.....WW.
W.W.W.....W.
.W.W......W.
..W.......W.

Sample Output

answer：

#include<stdio.h>
#include<string.h>
int n,m;
char s[104][104];
int dir[8][2]= {-1,-1,-1,0,-1,1,0,-1,0,1,1,-1,1,0,1,1};  // 二维数组，定义一个坐标的四周！八行二列。

bool vis[104][104];
void dfs(int x,int y)      //调用函数:DFS的精髓，也是最难理解的部分！

   {
    int i;
    if(x<0||x>=n||y<0||y>=m||vis[x][y]||s[x][y]=='.')    // 判定条件：池塘的边界！

      return;

          vis[x][y]=1;
    for(i=0; i<8; i++)
         dfs(x+dir[i][0],y+dir[i][1]);      //判断该坐标的四周，不满足if，就标记为1.  
              }
 int main()
{
    int i,j,ans;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(s,0,sizeof(s));
         memset(vis,0,sizeof(vis))；     //把所有数标记为零！表明未被、标记前为零

        ans=0;
        for(i=0; i<n; i++)
            scanf("%s",s[i]);
        for(i=0; i<n; i++)
            for(j=0; j<m; j++)
                if(s[i][j]=='W'&&!vis[i][j])      //判断是否标记并且是否为池塘！

                {
                    ans++;
                    dfs(i,j);
                }
        printf("%d\n",ans);
    }
    return 0;
}

#include<stdio.h>
#include<string.h>
int n,m;
char s[104][104];
int dir[8][2]= {-1,-1,-1,0,-1,1,0,-1,0,1,1,-1,1,0,1,1};
bool vis[104][104];
void dfs(int x,int y)
{
    int i;
    if(x<0||x>=n||y<0||y>=m||vis[x][y]||s[x][y]=='.') return;
    vis[x][y]=1;
    for(i=0; i<8; i++)
        dfs(x+dir[i][0],y+dir[i][1]);
}
int main()
{
    int i,j,ans;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(s,0,sizeof(s));
        memset(vis,0,sizeof(vis));
        ans=0;
        for(i=0; i<n; i++)
            scanf("%s",s[i]);
        for(i=0; i<n; i++)
            for(j=0; j<m; j++)
                if(s[i][j]=='W'&&!vis[i][j])
                {
                    ans++;
                    dfs(i,j);
                }
        printf("%d\n",ans);
    }
    return 0;
}