dfs入门

最新推荐文章于 2024-04-18 19:29:45 发布

JooooohnDoe

最新推荐文章于 2024-04-18 19:29:45 发布

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分类专栏：搜索之dfs与bfs

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本文介绍了一道经典的DFS算法入门题目，通过遍历二维矩阵来确定由相邻水域构成的湖泊总数。文章提供了完整的C++代码实现，并解释了如何通过深度优先搜索标记并计数这些湖泊。

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dfs经典入门题

Zesty Zapus

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

Hint

OUTPUT DETAILS:

There are three ponds: one in the upper left, one in the lower left,and one along the right side.

#include<stdio.h>
#include<string.h>

using namespace std;

char map[105][105];
int N,M;
int vis[8][2]={{-1,0},{-1,1},{0,1},{1,1},{1,0},{1,-1},{0,-1},{-1,-1}};

void dfs(int x, int y )
{
    for(int i=0;i<8;i++)
    {
        int x_next=x+vis[i][0];
        int y_next=y+vis[i][1];
        if(x_next>=0&&x_next<N  &&  y_next>=0&&y_next<M  &&  map[x_next][y_next]=='W')
        {
            map[x_next][y_next]='.';  //标记已经搜索过的点
            dfs(x_next,y_next);
        }
    }
    return ;
}


int main()
{
    int ans=0;
    memset(map,0,sizeof(map));
    scanf("%d%d",&N,&M);
    for(int i=0;i<N;i++)
        scanf("%s",map[i]);

    for(int i=0;i<N;i++)
        for(int j=0;j<M;j++)
        {
            if(map[i][j]=='W')
            {
                map[i][j]='.';
                dfs(i,j);
                ans++;
            }
        }

    printf("%d\n",ans);
    return 0;
}