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.
因为n最大为4，暴力搜索一下吧
#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;
int d[4][2] = { {0, 0}, {0, 1}, {1, 0}, {1, 1} };
int dir[3][3][2] = { { {0, 1}, {0, 0} }, { {1, 0}, {0, 0} }, { {0, 0} } };
int cnt[3] = {2, 2, 1};
int r, v[5][5], tmp;
const int c = 4;
bool ok(int k, int x, int y)
{
for (int i = 0; i< cnt[k]; i++)
{
int p = x + dir[k][i][0], q = y + dir[k][i][1];
if (p<= 0 || p >r) return false;
if (q<= 0 || q >c) return false;
if (v[p][q]) return false;
}
return true;
}
void clear(int k, int x, int y, int t)
{
for (int i = 0; i< cnt[k]; i++)
{
int p = x + dir[k][i][0], q = y + dir[k][i][1];
v[p][q] = t;
}
}
void dfs(int x, int y)
{
if (y >c) x = x + 1, y = 1;
if (x == r + 1)
{
tmp++;
return;
}
if (v[x][y]) dfs(x, y + 1);
for (int i = 0; i< 3; i++)
{
if (ok(i, x, y))
{
clear(i, x, y, 1);
dfs(x, y + 1);
clear(i, x, y, 0);
}
}
}
int find(int x, int y)
{
memset(v, 0, sizeof(v));
for (int i = 0; i< 4; i++)
v[x + d[i][0]][y + d[i][1]] = 1;
tmp = 0;
dfs(1, 1);
return tmp;
}
int solve()
{
int ans = 0;
for (int i = 1; i< r; i++)
{
for (int j = 1; j< c; j++)
{
ans += find(i, j);
}
}
return ans;
}
int main ()
{
int t[10];
for (r = 1; r<= 4; r++)
{
t[r] = solve();
}
int cas, n;
scanf("%d", &cas);
while (cas--)
{
scanf("%d", &n);
printf("%d\n", t[n]);
}
return 0;
}
原文地址：http://blog.csdn.net/libin56842/article/details/41847173
