fzu 2107 Hua Rong Dao(状态压缩)

最新推荐文章于 2018-04-13 14:53:11 发布

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最新推荐文章于 2018-04-13 14:53:11 发布

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本文详细探讨了胡同道上人数排列的算法问题，通过实例介绍了如何利用递归和状态转移矩阵解决类似问题。重点分析了曹操、交叉将军、垂直将军和士兵在特定矩形空间中的排列方式，提供了求解复杂排列问题的有效方法。

Problem 2107 Hua Rong Dao

Accept: 106 Submit: 197
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

212

Sample Output

018

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

#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
const int N=(1<<4);
int f[6][N][2]={0};//f[i][j][k]//放完第i-1行第i行的状态j，k=1放曹操
void dfs(int row,int col,int pre,int now,int cao,int k){
	if(col>=4){//放完4列,pre={1111}
		f[row][now][cao]+=k;//放完pre得到f[now]+=f[pre]
		//cout<<row<<" "<<now<<" "<<cao<<endl;
		return;
	}
	if(pre&(1<<col)){//第col已经放过
		dfs(row,col+1,pre,now,cao,k);
		return;
	}
	//a grid
	dfs(row,col+1,pre|(1<<col),now,cao,k);
	//a 1*2
	dfs(row,col+1,pre|(1<<col),now|(1<<col),cao,k);//放一竖，多出一块
	int t=(1<<col)|(1<<(col+1));
	if(col<3&&(pre&(1<<(col+1)))==0){
		//a 2*1
		dfs(row,col+1,pre|t,now,cao,k);
		//put caocao
		if(cao==0)dfs(row,col+1,pre|t,now|t,1,k);
	}
}
int main(){
	int i,j,k;
	f[0][N-1][0]=1;
	for(i=1;i<=5;i++)
		for(j=0;j<N;j++){
			if(f[i-1][j][0])dfs(i,0,j,0,0,f[i-1][j][0]);
			if(f[i-1][j][1])dfs(i,0,j,0,1,f[i-1][j][1]);
		}
	scanf("%d",&k);
	while(k--){
		scanf("%d",&i);
		printf("%d\n",f[i+1][0][1]);
	}
return 0;
}