POJ - 2386 Lake Counting

最新推荐文章于 2022-07-25 13:56:36 发布

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本文介绍了一个湖泊计数问题，通过深度优先搜索（DFS）算法来计算二维矩阵中湖泊的数量。湖泊由相邻的水域组成，而矩阵由干地和水域构成。文章提供了完整的代码实现，并附带了一个样例输入和输出。

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Lake Counting
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 39457 Accepted: 19558
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

3
Hint

OUTPUT DETAILS:

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

这里写代码片code：
#include<iostream>
using namespace std;

char sqares[105][105];
int n,m,ans;

void DFS(int x,int y);
int main(){
    ans = 0;
    cin>>n>>m;
    for(int i = 0;i < n;i++)
        for(int j = 0;j < m;j++)
            cin>>sqares[i][j];
    for(int i = 0;i < n;i++)
        for(int j = 0;j < m;j++)
            if(sqares[i][j]=='W'){
                DFS(i,j);
                ans++;
            }
    cout<<ans<<endl;
    return 0;

}
void DFS(int x,int y){
    sqares[x][y] = '.';
    for(int i = -1;i < 2;i++){
        for(int j = -1;j < 2;j++){
            ///判断是否在范围内以及该处是否有水如果符合条件就进行搜索
            if(x+i>=0&&x+i<=n-1&&y+j>=0&&y+j<=m-1&&sqares[x+i][y+j]=='W'){
                DFS(x+i,y+j);
            }
        }
    }
    return ;
}