ACM: 简单 动态规划题 poj 3036

                                      Honeycomb Walk

 

Description

A bee larva living in ahexagonal cell of a large honeycomb decides to creep for a walk. Ineach “step” the larva may move into any of the six adjacent cellsand after n steps, it is to end up in its originalcell.

Your program has to compute, for a given n, the number ofdifferent such larva walks.

Input

The first line contains aninteger giving the number of test cases to follow. Each caseconsists of one line containing an integer n, where 1 ≤n ≤ 14.

Output

For each test case, outputone line containing the number of walks. Under the assumption 1 ≤n ≤ 14, the answer will be less than2³¹.

Sample Input

2
2
4

Sample Output

6
90

 

题意: 蜜蜂可以在当前区域走向相邻的6个区域, 问你蜜蜂走n步可以回到原来的区域有多少种走法.

 

解题思路:

         1. 动态规划状态: dp[k][i][j]: 表示在到i,j区域的时候有多少种走法.(1<= i,j<= 16)

         2. 状态转移方程: dp[k][i][j] += dp[k-1][i-dir[t][0]][j-dir[t][1]];(0<t<6);

代码:

#include<cstdio>
#include <iostream>
#include <cstring>
using namespace std;
#define MAX 17

int dir[6][2] = {{1,0}, {-1,0}, {0,1}, {0,-1},{1,-1}, {-1,1}};

int n;
int dp[MAX][MAX][MAX];

void DP()
{
 memset(dp,0,sizeof(dp));
 dp[0][8][8] = 1;
 for(int k = 1; k <= 14; ++k)
 {
  for(int i = 1; i< MAX; ++i)
  {
   for(int j =1; j < MAX; ++j)
   {
    for(intt = 0; t < 6; ++t)
    {
     dp[k][i][j]+= dp[k-1][i-dir[t][0]][j-dir[t][1]];

    }
   }
  }
 }
}

int main()
{
// freopen("input.txt","r",stdin);
 DP();
 int caseNum;
 scanf("%d",&caseNum);
 while(caseNum--)
 {
  scanf("%d",&n);
  printf("%d\n",dp[n][8][8]);
 }

 return 0;
}