POJ 2386 DFS深搜入门

题目链接

Time Limit: 1000MS

Memory Limit: 65536K

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

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.

解题思路

简单的DFS入门，写好之后一直WA，然后发现scanf的一个注意事项之前一直没注意，cin读入字符会自动忽略换行符和空格，但是scanf不会，所以涉及到换行和空格的时候要用getchar()跳过换行符和空格。

AC代码

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

const int N = ;
char map[N][N];
int vis[N][N];//访问标记
int n, m;
int dx[] = { -,,,-,,-,, };
int dy[] = { ,,,,,-,-,- };//结点周边从上到下,从左到右八个点
int ans = ;//湖数

bool valid(int x, int y)
{
    return (x >=  && y >=  && x < n && y < m);
}

void DFS(int x, int y)
{
    vis[x][y] = ;//已访问
    for (int i = ; i < ; i++)
    {
        int newx = x + dx[i];
        int newy = y + dy[i];
        if (valid(newx, newy))
        {
            if (map[newx][newy] == 'W'&&vis[newx][newy] == -) DFS(newx, newy);
        }
    }
}

int main()
{
    memset(vis, -, sizeof(vis));
    scanf("%d%d", &n, &m);
    getchar();
    for (int i = ; i < n; i++)
    {
        for (int j = ; j < m; j++)
        {
            scanf("%c", &map[i][j]);
        }
        getchar();
    }
    for (int i = ; i < n; i++)
    {
        for (int j = ; j < m; j++)
        {
            if (map[i][j] == 'W'&&vis[i][j] == -)
            {
                ans++;
                DFS(i, j);
            }
        }
    }
    printf("%d", ans);
    return ;
}

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<queue>
using namespace std;

char map[][];
int vis[][] = {  };
int n, m;

int dx[] = { -,,,-,,-,, };
int dy[] = { ,,,,,-,-,- };

bool valid(int x, int y)
{
    return(x >=  && x < n&&y >=  && y < m);
}

void dfs(int x, int y)
{
    vis[x][y] = ;
    for (int i = ; i < ; i++)
    {
        int nx = x + dx[i];
        int ny = y + dy[i];
        if (valid(nx, ny))
        {
            if (map[nx][ny] == 'W' && !vis[nx][ny])dfs(nx, ny);
        }
    }
}

int main()
{
    scanf("%d%d", &n, &m);
    int cnt = ;
    for (int i = ; i < n; i++) scanf("%s", map[i]);
    for (int i = ; i < n; i++)
    {
        for (int j = ; j < m; j++)
        {
            if (!vis[i][j] && map[i][j] == 'W')
            {
                dfs(i, j);
                cnt++;
            }
        }
    }
    printf("%d\n", cnt);
    return ;
}

二刷

更简单的解决：不用vis数组，直接将访问过的“W”变成“.”。