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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
Sample Input
2 1 2
Sample Output
0 18
Hint
Here are 2 possible ways for the Hua Rong Dao 2*4.
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>
#include <iomanip>
#include <math.h>
#include <map>
using namespace std;
#define FIN freopen("input.txt","r",stdin);
#define FOUT freopen("output.txt","w",stdout);
#define INF 0x3f3f3f3f
#define INFLL 0x3f3f3f3f3f3f3f
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
typedef long long LL;
typedef pair<int,int> PII;
int n;
int ans, flag;
int vis[10][10];
bool check(int x, int y){
return x >= 1 && x <= n && y >= 1 && y <= 4 && vis[x][y] != 1;
}
void dfs(int cnt){
if(cnt == 4 * n && flag == 1){
ans ++;
return ;
}
if(cnt >= 4 * n) return ;
for(int i = 1; i <= n; i ++){
for(int j = 1; j <= 4; j ++){
if(check(i, j) && check(i + 1, j) && check(i, j + 1) && check(i + 1, j + 1) && !flag){
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, j) && check(i + 1, j)){
vis[i][j] = vis[i + 1][j] = 1;
dfs(cnt + 2);
vis[i][j] = vis[i + 1][j] = 0;
}
if(check(i, j) && check(i, j + 1)){
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()
{
//FIN
int T;
scanf("%d", &T);
while(T--){
memset(vis, 0, sizeof(vis));
scanf("%d", &n);
ans = 0;
flag = 0;
dfs(0);
printf("%d\n", ans);
}
}
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原文地址：http://www.cnblogs.com/Hyouka/p/5790889.html
