POJ 2386 （深搜或广搜）

Lake Counting

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 17917		Accepted: 9069

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

这道题目比较基础，可以深搜也可以广搜

深搜快一点 36ms，广搜47ms。很好的练习题目。

把注释的去掉就是深搜了。

#include
#include
using namespace std;
#define N 102

char map[N][N];
int n,m;
struct Node
{
    int x,y;
};
int dis[10][3]={ {0,1},{0,-1},{-1,0},{1,0},{-1,-1},{-1,1},{1,-1},{1,1} };
void search(int a,int b)
{
    /*map[a][b]='.';
    for(int i=0;i<8;i++)
    {
        int tmpx=a+dis[i][0];
        int tmpy=b+dis[i][1];
        if(tmpx>0 && tmpx<=n && tmpy>0 && tmpy<=m && map[tmpx][tmpy]=='W' ){
            //map[tmpx][tmpy]='.';
            search(tmpx,tmpy);
        }
    }*/
    Node tmp,be;
    be.x=a,be.y=b;
    queue v;
    v.push(be);
    map[a][b]='.';
    while(!v.empty())
    {
        be=v.front();
        v.pop();
        for(int i=0;i<8;i++)
        {
            tmp.x=be.x+dis[i][0];
            tmp.y=be.y+dis[i][1];
            if(tmp.x>0 && tmp.x<=n && tmp.y>0 && tmp.y<=m && map[tmp.x][tmp.y]=='W' )
            {
                v.push(tmp);
                map[tmp.x][tmp.y]='.';
            }
        }
    }
}

int main()
{
    int i,j,count;
    while(~scanf("%d %d",&n,&m))
    {
        getchar();
        count=0;
        for(i=1;i<=n;i++){
            for(j=1;j<=m;j++){
                scanf("%c",&map[i][j]);}  ///缺了括号，害死人啊
            getchar();}
        for(i=1;i<=n;i++)
            for(j=1;j<=m;j++)
                if(map[i][j]=='W')
                {
                    //printf("%d %d\n",i,j);
                    search(i,j);count++;
                }
        printf("%d\n",count);
    }
    return 0;
}

[cpp]   view plain  copy

[cpp]   view plain  c

[cpp]   view plain  cop

[cpp]   view plain  copy

#include
#include
using namespace std;
#define N 102

char map[N][N];
int n,m;

void search(int i,int j)
{
    if(map[i][j-1]=='W') { map[i][j-1]='.'; search(i,j-1); }
    if(map[i][j+1]=='W') { map[i][j+1]='.'; search(i,j+1); }
    if(map[i-1][j]=='W') { map[i-1][j]='.'; search(i-1,j); }
    if(map[i+1][j]=='W') { map[i+1][j]='.'; search(i+1,j); }
    if(map[i-1][j-1]=='W') { map[i-1][j-1]='.'; search(i-1,j-1); }
    if(map[i-1][j+1]=='W') { map[i-1][j+1]='.'; search(i-1,j+1); }
    if(map[i+1][j-1]=='W') { map[i+1][j-1]='.'; search(i+1,j-1); }
    if(map[i+1][j+1]=='W') { map[i+1][j+1]='.'; search(i+1,j+1); }
}

int main()
{
    int i,j,count;
    while(~scanf("%d %d",&n,&m))
    {
        getchar();
        count=0;
        for(i=1;i<=n;i++){
            for(j=1;j<=m;j++){
                scanf("%c",&map[i][j]);}  ///缺了括号，害死人啊
            getchar();}
        for(i=1;i<=n;i++)
            for(j=1;j<=m;j++)
                if(map[i][j]=='W')
                {
                    //printf("%d %d\n",i,j);
                    search(i,j);count++;
                }
        printf("%d\n",count);
    }
    return 0;
}
题意：有一个大小为N×M的园子，雨后积起了水。八连通的积水被认为是连在一起的。求园子里一共有多少水洼？* * ** W*    （八连通指的就是左图中相对W的*的部分）* * *Sample Input10 12
W........WW.
.WWW.....WWW
....WW...WW.
.........WW.
.........W..
..W......W..
.W.W.....WW.
W.W.W.....W.
.W.W......W.
..W.......W.Sample Output3
PS:样例中的3个水洼分别是在左上，左下，和右方。
这是的一道入门的深度优先搜索题，很简单~~
从任意的W开始，不停地把邻接的部分（八连通）用'.'代替。1次DFS后与初始的这个W连接的所有W就都被替换成了'.'，因此直到图中不在存在W为止，总共进行DFS的次数就是答案了。8个方向的共对应了8种状态转移，每个格子作为DFS的参数至多被调用一次，所以复杂度为O(8×M×N）=O(M×N)。
 
      
       
        
       
      #include
#include
using namespace std;
char field[105][105];
int N,M;
void dfs(int x, int y)
{
    //现在的位置（x,y)
    field[x][y] = '.';//将现在所在位置替换为'.'
    for (int dx = -1; dx <= 1; dx++)//循环遍历连通的8个方向
    {
        for (int dy = -1; dy <= 1; dy++)
        {
            int nx = x + dx, ny = y + dy;//向x方向移动dx,向y方向移动dy,移动的结果为(nx,ny)
            if (0 <= nx&&nx <= N && 0 <= ny&&ny <= M&&field[nx][ny] == 'W')//判断(nx,ny)是不是在园子里，以及是否有积水
                dfs(nx, ny);
        }
    }
}
int main()
{
    cin >> N >> M;
    for (int i = 0; i < N; i++)
        for (int j = 0; j < M; j++)
            cin >> field[i][j];
    int res = 0;
    for(int i=0;i)
        for(int j=0;j)
            if (field[i][j] == 'W')//从有积水的地方开始dfs
            {
                dfs(i, j);
                res++;
            }
    cout << res << endl;
    return 0;
} 
      
     
       HDU1241&POJ2386 dfs简单题 
      
      
      2道题目都差不多，就是问和相邻所有点都有相同数据相连的作为一个联通快，问有多少个连通块 
        
      因为最近对搜索题目很是畏惧，总是需要看别人代码才能上手，就先拿这两道简单的dfs题目来练练手，顺便理一理dfs的思路，分析清楚dfs的退出递归的条件和什么时候进行递归调用是至关重要的，这两道题目不涉及回溯，对于需要回溯的题目也要清楚分析，找到回溯条件，在对一个新的状态dfs时，后面加上回溯的语句 
        
      HDU1241代码: 
       
        
         
        
        1 #include 
 2 #include 
 3 #include 
 4 using namespace std;
 5 
 6 int n,m,ans;
 7 int visit[102][102],dir[8][2] = {{1,0},{-1,0},{0,1},{0,-1},{1,1},{1,-1},{-1,1},{-1,-1}};
 8 char mat[102][102];
 9 
10 void dfs(int a,int b)
11 {
12     visit[a][b] = 1;
13     for(int i=0;i<8;i++){
14         int xx = a+dir[i][0];
15         int yy = b+dir[i][1];
16         if(xx>=0&&xx=0&&yy'@'){
17             if(!visit[xx][yy])
18                 dfs(xx,yy);
19         }
20     }
21 }
22 int main()
23 {
24     while(~scanf("%d%d",&n,&m)){
25         if(n==0&&m==0) break;
26 
27         for(int i=0;i){
28             for(int j=0;j)
29                 cin>>mat[i][j];
30         }
31 
32         memset(visit,0,sizeof(visit));
33         ans = 0;
34 
35         for(int i=0;i)
36         {
37             for(int j=0;j)
38                 if(!visit[i][j]&&mat[i][j] == '@')
39                 {
40                     dfs(i,j);
41                     ans++;
42                 }
43         }
44 
45         printf("%d\n",ans);
46     }
47     return 0;
48 } 
        
         
        
       
        
      POJ2386代码: 
       
        
         
        
        1 #include 
 2 #include 
 3 
 4 using namespace std;
 5 
 6 #define N 102
 7 
 8 int n,m,ans;
 9 int visit[N][N],dir[8][2] = {{1,0},{-1,0},{0,1},{0,-1},{1,1},{1,-1},{-1,1},{-1,-1}};
10 char mat[N][N];
11 
12 void dfs(int x,int y)
13 {
14     visit[x][y]=1;
15     for(int i=0;i<8;i++){
16         int xx=x+dir[i][0];
17         int yy=y+dir[i][1];
18         if(xx>=0&&x=0&&yy'W'&&!visit[xx][yy])
19             dfs(xx,yy);
20     }
21 }
22 
23 int main()
24 {
25     while(~scanf("%d%d",&n,&m)){
26         for(int i=0;i)
27             scanf("%s",mat[i]);
28 
29         memset(visit,0,sizeof(visit));
30         ans=0;
31 
32         for(int i = 0 ;i)
33         for(int j=0;j){
34             if(mat[i][j] == 'W'&&!visit[i][j])
35             {
36                 dfs(i,j);
37                 ans++;
38             }
39         }
40 
41         printf("%d\n",ans);
42     }
43     return 0;
44 }