FZU 2107 Hua Rong Dao——————DFS

港控/mmm° 2022-02-21 10:59 48阅读 0赞

# Hua Rong Dao #

Accept: 523 Submit: 1086  
Time Limit: 1000 mSec Memory Limit : 32768 KB  
**Problem Description**  
Cao Cao was hunted down by thousands of enemy soldiers when he escaped from Hua Rong Dao. Assuming Hua Rong Dao is a narrow aisle (one N*4 rectangle), while Cao Cao can be regarded as one 2*2 grid. Cross general can be regarded as one 1*2 grid.Vertical general can be regarded as one 2*1 grid. Soldiers can be regarded as one 1\*1 grid. Now Hua Rong Dao is full of people, no grid is empty.

There is only one Cao Cao. The number of Cross general, vertical general, and soldier is not fixed. How many ways can all the people stand?

**Input**  
There is a single integer T (T≤4) in the first line of the test data indicating that there are T test cases.

Then for each case, only one integer N (1≤N≤4) in a single line indicates the length of Hua Rong Dao.

**Output**  
For each test case, print the number of ways all the people can stand in a single line.  
**Sample Input**  
2  
1  
2  
**Sample Output**  
0  
18  
**Hint**  
Here are 2 possible ways for the Hua Rong Dao 2\*4.  
![在这里插入图片描述][10006_codeispoetry.jpg]

**Source**  
“高教社杯”第三届福建省大学生程序设计竞赛

--------------------

简单DFS，需要将搜索的顺序方向和大小自己设定  
可是做题的时候还是没写出来(难受

// #include<bits/stdc++.h>
    #include<cstdio>
    #include<cstring>
    #include<string>
    #include<iostream>
    using namespace std;
    int n,ans,flag;
    int vis[10][10];
    
    bool check(int x,int y)
    {
        
            return x>=1 && y>=1 && x<=n && y<=4 && vis[x][y]!=1;
    }
    
    void dfs(int cnt)
    {
        
            if(cnt==n*4&&flag==1)
            {
        
                    ans ++;
                    return ;
            }
            if(cnt>=n*4)    return ;
            for(int i=1;i<=n;++i)
                    for(int j=1;j<=4;++j)//对于任意一点，都有四个选择
                    {
        
                            if(check(i,j)&&check(i+1,j)&&check(i,j+1)&&check(i+1,j+1)&&!flag)//放曹操，如果曹操还没有放下的话
                            {
        
                                    vis[i][j]=vis[i+1][j]=vis[i][j+1]=vis[i+1][j+1]=1;
                                    flag = 1;
                                    dfs(cnt+4);
                                    vis[i][j]=vis[i+1][j]=vis[i][j+1]=vis[i+1][j+1]=0;
                                    flag = 0;
                            }
                            if(check(i+1,j)&&check(i,j))
                            {
        
                                    vis[i][j] = vis[i+1][j] = 1;
                                    dfs(cnt+2);
                                    vis[i][j] = vis[i+1][j] = 0;
                            }
                            if(check(i,j+1)&&check(i,j))
                            {
        
                                    vis[i][j] = vis[i][j+1] = 1;
                                    dfs(cnt+2);
                                    vis[i][j] = vis[i][j+1] = 0;
                            }
                            if(check(i,j))
                            {
        
                                    vis[i][j] = 1;
                                    dfs(cnt+1);
                                    vis[i][j] = 0;
                                    return ;
    
                            }
                    }
    
    }
    
    int main()
    {
        
            int t;
            cin>>t;
            while(t--)
            {
        
                    cin>>n;
                    ans = 0;
                    flag = 0;
                    dfs(0);
                    cout<<ans<<endl;
            }
            return 0;
    }

[10006_codeispoetry.jpg]: http://acm.fzu.edu.cn/image/Problem/10006_codeispoetry.jpg