FZU 2107 Hua Rong Dao 递归回溯

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最新推荐文章于 2025-05-05 22:22:17 发布

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本文探讨了经典的华容道问题，在一个N*4的矩形中放置2*2、1*2、2*1及1*1大小的矩形，确保至少包含一个2*2的矩形，通过递归枚举所有可能的放置方式来计算解决方案的数量。

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Problem 2107 Hua Rong Dao

Accept: 254 Submit: 591
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

212

Sample Output

018

Hint

Here are 2 possible ways for the Hua Rong Dao 2*4.

Source

“高教社杯”第三届福建省大学生程序设计竞赛

在n*4的矩形里面放满一些小的矩形，要求必须有一个2*2的矩形，还有1*2,2*1,1*1的矩形，求有多少种方法满足条件。

递归枚举所有情况。

//15 ms	208KB
#include<stdio.h>
#include<string.h>
int vis[10][10];
int n,ans,flag;
bool judge(int x,int y)
{
    if(x>=1&&x<=n&&y>=1&&y<=4&&!vis[x][y])return true;
    return false;
}
void dfs(int count)
{
    if(count==n*4&&flag){ans++;return;}
    if(count>=4*n)return;
    for(int i=1;i<=n;i++)
        for(int j=1;j<=4;j++)
        {
            if(judge(i,j)&&judge(i,j+1)&&judge(i+1,j)&&judge(i+1,j+1)&&!flag)
            {
                flag=1;
                vis[i][j]=vis[i][j+1]=vis[i+1][j]=vis[i+1][j+1]=1;
                dfs(count+4);
                flag=0;
                vis[i][j]=vis[i][j+1]=vis[i+1][j]=vis[i+1][j+1]=0;
            }
            if(judge(i,j)&&judge(i,j+1))
            {
                vis[i][j]=vis[i][j+1]=1;
                dfs(count+2);
                vis[i][j]=vis[i][j+1]=0;
            }
            if(judge(i,j)&&judge(i+1,j))
            {
                vis[i][j]=vis[i+1][j]=1;
                dfs(count+2);
                vis[i][j]=vis[i+1][j]=0;
            }
            if(judge(i,j))
            {
                vis[i][j]=1;
                dfs(count+1);
                vis[i][j]=0;
                return;
            }
        }
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        ans=flag=0;
        memset(vis,0,sizeof(vis));
        scanf("%d",&n);
        dfs(0);
        printf("%d\n",ans);
    }
    return 0;
}