hdu Lake Counting

最新推荐文章于 2019-08-12 16:47:06 发布

hn_Lin

最新推荐文章于 2019-08-12 16:47:06 发布

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分类专栏： bfs dfs

本文链接：https://blog.youkuaiyun.com/weixin_41760511/article/details/81269028

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本文介绍了一种使用深度优先搜索(DFS)算法解决计数水塘问题的方法。该问题要求在给定的矩阵中找出所有由'W'(水)组成的连通区域的数量。文章通过示例解释了输入输出格式，并提供了一个简洁的C++代码实现。

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Lake Counting

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 131072/65536K (Java/Other)

Problem 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< br>< br>* 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

Sample Output

题意： w就是水坑，边上八个位置有w的话就算他们是相连的

求总共有多少个水坑。很直接的dfs算法。

#include <iostream>
#include <cstring>
using namespace std;

char map1[110][110];
int f[110][110];

void dfs(int x, int y)
{
    if(f[x][y]==1||map1[x][y]=='.'||map1[x][y]==0)
        return;
    f[x][y] = 1;
    dfs(x-1,y-1);
    dfs(x-1,y);
    dfs(x-1,y+1);
    dfs(x,y-1);
    dfs(x,y+1);
    dfs(x+1,y-1);
    dfs(x+1,y);
    dfs(x+1,y+1);
}

int main()
{
    memset(map1,0,sizeof(map1));
    memset(f,0,sizeof(f));
    int m,n,con = 0;
    cin >> m >> n;
    for(int i = 1; i<=m; ++i)
        for(int j = 1; j<=n; ++j)
            cin >> map1[i][j];
    for(int i = 1; i <= m; ++i)
        for(int j = 1; j <= n; ++j)
        {
            if(!f[i][j] && map1[i][j] == 'W')
            {
                con++;
                dfs(i,j);
            }
        }
    cout << con<<endl;
    return 0;
}