POJ - 3797 Tiling a Grid With Dominoes（递推）

Tiling a Grid With Dominoes

Description

We wish to tile a grid 4 units high and N units long with rectangles (dominoes) 2 units by one unit (in either orientation). For example, the figure shows the five different ways that a grid 4 units high and 2 units wide may be tiled. 



Write a program that takes as input the width, W, of the grid and outputs the number of different ways to tile a 4-by-W grid.
Input

The first line of input contains a single integer N, (1 ≤ N ≤ 1000) which is the number of datasets that follow. 

Each dataset contains a single decimal integer, the width, W, of the grid for this problem instance.
Output

For each problem instance, there is one line of output: The problem instance number as a decimal integer (start counting at one), a single space and the number of tilings of a 4-by-W grid. The values of W will be chosen so the count will fit in a 32-bit integer.
Sample Input

3
2
3
7

Sample Output

1 5
2 11
3 781

Source

Greater New York Regional 2007

题意：给你无数个1*2的方块，问你拼出4*N长度的矩形有多少种拼法。

解题思路：递推即可，详细看图片和代码。这道题其实是用来练习状压DP的，但是不会写……以后学了插头DP再回来看看。



          图0

     

          

      


         图  1                                                                                图2

#include<iostream>
#include<deque>
#include<memory.h>
#include<stdio.h>
#include<map>
#include<string>
#include<algorithm>
#include<vector>
#include<math.h>
#include<stack>
#include<queue>
#include<set>
#define INF 1<<29
using namespace std;

//图0
int dp[1005];
int a[1005];
int b[1005];
int c[1005];

int main()
{
dp[0]=1;dp[1]=1;
a[0]=0;a[1]=1;
b[0]=1;b[1]=1;
c[0]=1;c[1]=1;

for(int i=2;i<1005;i++){
dp[i]=a[i-1]+b[i-1]+c[i-1]+dp[i-1]+dp[i-2];//图1
a[i]=a[i-2]+dp[i-1];//图2

b[i]=c[i-1]+dp[i-1];//自己思考
c[i]=b[i-1]+dp[i-1];//自己思考
}

int t;
scanf("%d",&t);
for(int qqq=1;qqq<=t;qqq++){
int w;
scanf("%d",&w);
printf("%d %d\n",qqq,dp[w]);

}

return 0;
}