foj Problem 2107 Hua Rong Dao

Problem 2107 Hua Rong Dao

Accept: 503    Submit: 1054
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.

Source

“高教社杯”第三届福建省大学生程序设计竞赛

 

题意：搜索多少种布阵方式，一定要有曹操。

思路：回溯dfs。

AC代码：

#define _CRT_SECURE_NO_DEPRECATE
#include <iostream>
#include<vector>
#include<algorithm>
#include<cstring>
#include<bitset>
#include<set>
#include<map>
#include<cmath>
#include<queue>
using namespace std;
#define N_MAX 7
#define MOD 1000000007
#define INF 0x3f3f3f3f
typedef long long ll;
int n, k;
bool vis[N_MAX][N_MAX],cc;
int ans = ;
void dfs(int x,int y) {
    if (x==n) {//搜索结束
        if (cc)ans++;//有曹操
        return;
    }
    int xx=x, yy=y+;//下次要去的点
    if (yy == ) {
        xx++, yy = ;
    }

    if (vis[x][y])dfs(xx, yy);
    else {
        for (int cs = ; cs < ;cs++) {
            if (cs == &&!cc) {
                if (x +  < n&&y +  <  && !vis[x][y] && !vis[x + ][y] && !vis[x][y + ] && !vis[x + ][y + ]) {
                    cc = true;
                    vis[x][y] = vis[x + ][y] = vis[x][y + ] = vis[x + ][y + ] = true;
                    dfs(xx,yy);
                    vis[x][y] = vis[x + ][y] = vis[x][y + ] = vis[x + ][y + ] = false;
                    cc = false;
                }
            }
            if (cs == ) {
                if (x +  < n && !vis[x][y] && !vis[x + ][y]) {
                    vis[x][y] = vis[x + ][y] = true;
                    dfs(xx,yy);
                    vis[x][y] = vis[x + ][y] = false;
                }
            }
            if (cs == ) {
                if (y +  <  && !vis[x][y] && !vis[x][y + ]) {
                    vis[x][y] = vis[x][y + ] = true;
                    dfs(xx, yy);
                    vis[x][y] = vis[x][y + ] = false;
                }
            }
            if (cs == ) {
                if (!vis[x][y]) {
                    vis[x][y] = true;
                    dfs(xx, yy);
                    vis[x][y] = false;
                }
            }
        }
    }
}

int main() {
    int t; cin >> t;
    while(t--){
        memset(vis, , sizeof(vis)); cc = ; ans = ;
        cin >> n;
        dfs(, );
    cout << ans<<endl;
    }
    return ;
}