Problem DescriptionCao 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?
InputThere 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 Sample Input Sample Output HintHere are 2 possible ways for the Hua Rong Dao 2*4.
Source#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 = 0; 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+1;//下次要去的点 if (yy == 4) { xx++, yy = 0; } if (vis[x][y])dfs(xx, yy); else { for (int cs = 0; cs < 4;cs++) { if (cs == 0&&!cc) { if (x + 1 < n&&y + 1 < 4 && !vis[x][y] && !vis[x + 1][y] && !vis[x][y + 1] && !vis[x + 1][y + 1]) { cc = true; vis[x][y] = vis[x + 1][y] = vis[x][y + 1] = vis[x + 1][y + 1] = true; dfs(xx,yy); vis[x][y] = vis[x + 1][y] = vis[x][y + 1] = vis[x + 1][y + 1] = false; cc = false; } } if (cs == 1) { if (x + 1 < n && !vis[x][y] && !vis[x + 1][y]) { vis[x][y] = vis[x + 1][y] = true; dfs(xx,yy); vis[x][y] = vis[x + 1][y] = false; } } if (cs == 2) { if (y + 1 < 4 && !vis[x][y] && !vis[x][y + 1]) { vis[x][y] = vis[x][y + 1] = true; dfs(xx, yy); vis[x][y] = vis[x][y + 1] = false; } } if (cs == 3) { 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, 0, sizeof(vis)); cc = 0; ans = 0; cin >> n; dfs(0, 0); cout << ans<<endl; } return 0; }
原文：https://www.cnblogs.com/ZefengYao/p/9158821.html