FZU 2107 Hua Rong Dao(dfs)
Problem 2107 Hua Rong Dao Accept: 318 Submit: 703 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
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#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <math.h>
#include <stdio.h>
using namespace std;
int vis[10][10];
int n;
int ans;
bool flag;
int judge(int x,int y)
{
if(x<1||x>n||y<1||y>4||vis[x][y])
return 0;
return 1;
}
void dfs(int count)
{
if(count==n*4&&flag)
{
ans++;
return;
}
if(count>=n*4)
return;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=4;j++)
{
if(judge(i,j)&&judge(i+1,j+1)&&judge(i,j+1)&&judge(i+1,j)&&!flag)
{
flag=1;
vis[i][j]=1;vis[i+1][j+1]=1;vis[i][j+1]=1;vis[i+1][j]=1;
dfs(count+4);
flag=0;
vis[i][j]=0;vis[i+1][j+1]=0;vis[i][j+1]=0;vis[i+1][j]=0;
}
if(judge(i,j)&&judge(i,j+1))
{
vis[i][j]=1;vis[i][j+1]=1;
dfs(count+2);
vis[i][j]=0;vis[i][j+1]=0;
}
if(judge(i,j)&&judge(i+1,j))
{
vis[i][j]=1;vis[i+1][j]=1;
dfs(count+2);
vis[i][j]=0;vis[i+1][j]=0;
}
if(judge(i,j))
{
vis[i][j]=1;
dfs(count+1);
vis[i][j]=0;
return;
}
}
}
}
int main()
{
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
memset(vis,0,sizeof(vis));
ans=0;
flag=0;
dfs(0);
printf("%d\n",ans);
}
return 0;
}腾讯云开发者
