南邮 OJ 1498 Honeycomb Walk

Honeycomb Walk

时间限制(普通/Java) :  1000 MS/ 3000 MS          运行内存限制 : 65536 KByte
总提交 : 34            测试通过 : 14 

比赛描述

    A bee larva living in a hexagonal cell of a large honeycomb decides to creep for a walk. In each “step” the larva may move into any of the six adjacent cells and after n steps, it is to end up in its original cell.
Your program has to compute, for a given n, the number of different such larva walks.

输入

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

输出

    For each test case, output one line containing the number of walks. Under the assumption 1 ≤ n ≤ 14, the answer will be less than 231 - 1.

样例输入

2
2
4

样例输出

6
90

题目来源

Nordic 2006

//dp[i][j][k] = 第 k 步走到(i,j)的走法
#include<iostream>
#define N 15
int dp[N][N][N];
int dirX[6] = { 1, 0,-1,-1, 0, 1};
int dirY[6] = { 0, 1, 1, 0,-1,-1};

int main(){
	int i,j,k,d,x,y;
	dp[7][7][0] = 1;
	for(k=1;k<N;k++){
		for(i=0;i<N;i++){
			for(j=0;j<N;j++){
				for(d=0;d<6;d++){
					x = i+dirX[d];
					y = j+dirY[d];
					if(x>=0 && x<N && y>=0 && y<N){
						dp[i][j][k] += dp[x][y][k-1];
					}
				}
			}
		}
	}
	int t,n;
	scanf("%d",&t);
	while(t--){
		scanf("%d",&n);
		printf("%d\n",dp[7][7][n]);
	}
}