搜索算法集锦 Ⅰ

女爷i 2022-08-03 11:55 200阅读 0赞

搜索：带有方向性的枚举（自我理解）  
分类：广度优先搜索(BFS)和深度优先搜索(DFS)（含有回溯）

\[hdu1312：\] ([http://acm.hdu.edu.cn/showproblem.php?pid=1312][http_acm.hdu.edu.cn_showproblem.php_pid_1312])  
题目描述：  
Red and Black

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)  
Total Submission(s): 13171 Accepted Submission(s): 8163

Problem Description  
There is a rectangular（矩形） room, covered with square tiles. Each tile（地砖） is colored either red or black. A man is standing on a black tile. From a tile, he can move to one of four adjacent（临近的） tiles. But he can’t move on red tiles, he can move only on black tiles.

Write a program to count the number of black tiles which he can reach by repeating the moves described above.

Input  
The input consists of multiple data sets. A data set starts with a line containing two positive integers W and H; W and H are the numbers of tiles in the x- and y- directions, respectively. W and H are not more than 20.

There are H more lines in the data set, each of which includes W characters. Each character represents the color of a tile as follows.

‘.’ - a black tile  
‘\#’ - a red tile  
‘@’ - a man on a black tile(appears exactly once in a data set)

Output  
For each data set, your program should output a line which contains the number of tiles he can reach from the initial（最初的） tile (including itself).

Sample Input  
6 9  
….\#.  
…..\#  
……  
……  
……  
……  
……

# @… #

.\#..\#.  
11 9  
.\#………  
.\#.\#\#\#\#\#\#\#.  
.\#.\#…..\#.  
.\#.\#.\#\#\#.\#.  
.\#.\#..@\#.\#.  
.\#.\#\#\#\#\#.\#.  
.\#…….\#.  
.\#\#\#\#\#\#\#\#\#.  
………..  
11 6  
..\#..\#..\#..  
..\#..\#..\#..  
..\#..\#..\#\#\#  
..\#..\#..\#@.  
..\#..\#..\#..  
..\#..\#..\#..  
7 7  
..\#.\#..  
..\#.\#..

### . ###

…@…

### . ###

..\#.\#..  
..\#.\#..  
0 0

Sample Output  
45  
59  
6  
13

*【言简意赅】：*  
计算从’@’开始能经过的所有’.’的数量，路径可重复，但总数不重复记录

广搜代码实现：

#include <iostream>
    #include <string.h>
    #include <stdio.h>
    using namespace std;
    
    const int maxn=30;
    char a[maxn][maxn];//存图数组
    bool b[maxn][maxn];//判重数组
    int qx[maxn*maxn],qy[maxn*maxn];
    int vb[4][2]= {
       {-1,0},{
       1,0},{
       0,1},{
       0,-1}};//上下左右四个方向
    int ans,m,n;
    void bfs(int x,int y)
    {
        int l=0,r=0;
        qx[r]=x;
        qy[r++]=y;
        //qx[r]=x;qy[r]=y;r++;
        b[x][y]=1;
        ans++;
        while(l<r)
        {
            int cx=qx[l];
            int cy=qy[l++];
            //int cx=qx[l];int cy=qy[l];l++;
            for(int i=0; i<4; i++)
            {
                int nx=cx+vb[i][0];
                int ny=cy+vb[i][1];
                if(nx>=1&&nx<=n&&ny>=1&&ny<=m&&!b[nx][ny]&&a[nx][ny]!='#')
                {
                    ans++;
                    b[nx][ny]=1;//状态标记
                    qx[r]=nx;qy[r++]=ny;//注意！此处是状态转移
                    //qx[r]=nx;qy[r]=ny;r++;
                }
            }
        }
    }
    int main()
    {
        int x,y;
        while(scanf("%d%d",&m,&n)!=EOF)//此题是应用广搜
        {
            getchar();
            if(m==0&&n==0)
            break;
            memset(a,0,sizeof(a));
            memset(b,false,sizeof(b));
            for(int i=1; i<=n; i++)
            {
                for(int j=1; j<=m; j++)
                {
                    cin>>a[i][j];
                    if(a[i][j]=='@')//由于只有一个'@'，即可以在输入时就将其下标记录下来
                    {
                        x=i;
                        y=j;
                    }
                }
            }
            ans=0;
            bfs(x,y);
            cout<<ans<<endl;
        }
        return 0;
    }

深搜代码实现：

#include <iostream>
    #include <cstdio>
    #include <cstring>
    
    using namespace std;
    
    const int maxn=30;
    int n,m;
    int dir[4][2]= {
       {
       1,0},{-1,0},{
       0,1},{
       0,-1}};
    char map[maxn][maxn];
    bool vis[maxn][maxn];
    int ans;
    void dfs(int x,int y)
    {
        vis[x][y]++;
        ans++;
        for(int i=0; i<4; i++)//在可选的4个方向上依次实现搜索
        {
            int nx=x+dir[i][0];
            int ny=y+dir[i][1];
            if(nx<1||nx>n) continue;
            if(ny<1||ny>m) continue;
            if(vis[nx][ny]||map[nx][ny]=='#') continue;
            dfs(nx,ny);//递归实现深度优先搜索
        }
    }
    int main()
    {
        int ca,a,b,cas=1;
        cin>>ca;
        while(ca--)
        {
            cin>>m>>n;
            int sx,sy;
            for(int i=1; i<=n; i++)
                for(int j=1; j<=m; j++)
                {
                    cin>>map[i][j];
                    if(map[i][j]=='@') sx=i,sy=j;
                }
            memset(vis,0,sizeof(vis));
            ans=0;
            dfs(sx,sy);
            cout<<"Case "<<cas++<<": "<<ans<<endl;
        }
        return 0;
    }

[http_acm.hdu.edu.cn_showproblem.php_pid_1312]: http://acm.hdu.edu.cn/showproblem.php?pid=1312