Hua Rong Dao（DFS）

Problem 2107 Hua Rong Dao

Accept: 183    Submit: 465
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.

      题意：

      给出 T，代表有 T 组数组，后给出 N（1 ~ 4）。代表有 N * 4 个格子，现有 4 种格子块去填充这 N * 4 个格子，而 CaoCao 是必须用的，但是只能用 1 个。问一共有多少种填充方式。

 

      思路：

      DFS。N 最大也只有4，所以不会超时。

 

      AC：

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

int vis[5][5], sum;
int n, way;

void dfs(int x, int y, int ans) {
        if (ans == sum) {
                ++way;
                return;
        }

        if (vis[x][y]) {
                if (y + 1 <= 4) dfs(x, y + 1, ans);
                else if (x + 1 <= n) dfs(x + 1, 1, ans);
        } else {

                vis[x][y] = 2;
                dfs(1, 1, ans + 1);
                vis[x][y] = 0;

                if (y + 1 <= 4 && !vis[x][y + 1]) {
                        vis[x][y] = vis[x][y + 1] = 3;
                        dfs(1, 1, ans + 2);
                        vis[x][y] = vis[x][y + 1] = 0;
                }

                if (x + 1 <= n && !vis[x + 1][y]) {
                        vis[x][y] = vis[x + 1][y] = 4;
                        dfs(1, 1, ans + 2);
                        vis[x][y] = vis[x + 1][y] = 0;
                }
        }

}

int main() {
        int t;
        scanf("%d", &t);
        while (t--) {

                scanf("%d", &n);

                sum = 4 * n;
                way = 0;

                for (int i = 1; i <= n - 1; ++i) {
                        for (int j = 1; j <= 3; ++j) {
                                memset(vis, 0, sizeof(vis));

                                vis[i][j] = 1;
                                vis[i][j + 1] = 1;
                                vis[i + 1][j] = 1;
                                vis[i + 1][j + 1] = 1;

                                dfs(1, 1, 4);
                        }
                }

                printf("%d\n", way);

        }
        return 0;
}