POJ-2386(深广搜基础)

Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 25322		Accepted: 12759

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

---

思路：
很基础的一道题目，入门级别，有助于理解搜索的本质以及bfs和dfs的区别
这个题可以用“扫雷”的思维来形象的理解，即遇到一个'W'，点一下这个点，则与它相邻的一片为‘W’的点都变成‘.’了
而我们只需要从头开始遍历一共有多少个这样的W即可

---

dfs：

#include <stdio.h>
#define maxn 107

int n,m;
char g[maxn][maxn];
int dir[10][2] = {{-1,-1},{-1,0},{-1,1},{0,-1},{0,1},{1,0},{1,-1},{1,1}};

void dfs(int x,int y){    
    g[x][y] = '.';//此行代码意义重大，相当于将其置为已访问状态    
    for(int i = 0;i < 8;i++) {        
        int dx = x+dir[i][0];        
        int dy = y+dir[i][1];        
        if(dx>n||dx<1||dy<1||dy>m)            
            continue;        
        if(g[dx][dy] == '.')            
            continue;        
        dfs(dx,dy);    
    }
}
int main()
{    
    while(scanf("%d%d",&n,&m)!=EOF)    {        
        int ans = 0;        
        for(int i = 1;i <= n;i++)            
            scanf("%s",g[i]+1);  
        int cnt = 0;
        for(int i = 1;i <= n;i++)        
        for(int j = 1;j <= m;j++) {
            if(g[i][j] == 'W') {                
                ans++;                
                dfs(i,j);//把所有和该点相邻的W都变成.            
            }        
        }
        printf("%d
",ans);    
    }    
    return 0;
}

bfs:

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

int n,m;
char G[107][107];
int dir[8][2] = {{-1,-1},{-1,0},{-1,1},{0,-1},{0,1},{1,-1},{1,0},{1,1}};
typedef pair<int,int> node;

void bfs(int x,int y)
{
    G[x][y] ='.';//比较好的一种处理方法，省去开vis数组
    priority_queue<node> q;//q中存储了（x，y）点的所有连通点 
    q.push(make_pair(x,y));
    while(!q.empty())
    {
        node t = q.top();
        q.pop();
        int tx = t.first;
        int ty = t.second; 
        for(int i = 0;i < 8;i++)
        {
            int dx = tx+dir[i][0];
            int dy = ty+dir[i][1];
            if(dx>=1&&dx<=n&&dy>=1&&dy<=m && G[dx][dy]=='W') {
                G[dx][dy] = '.';
                q.push(make_pair(dx,dy));
            }
        }
    }
}

int main()
{
    while(cin>>n>>m)
    {
        int ans = 0;
        for(int i = 1;i <= n;i++)
            scanf("%s",&G[i][1]);

        for(int i = 1;i <= n;i++)
            for(int j = 1;j <= m;j++) 
                if(G[i][j]=='W') {
                    bfs(i,j);
                    ans++;
                }
        cout<<ans<<endl;
    }
    return 0;
}