Honeycomb Walk
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 111 Accepted Submission(s): 69Problem Description 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.
Input 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.
Output For each test case, output one line containing the number of walks. Under the assumption 1 ≤ n ≤ 14, the answer will be less than 2 31< http://www.2cto.com/kf/ware/vc/" target="_blank" class="keylink">vc3VwPiAtIDEuPGJyPgoKIAo8YnI+ClNhbXBsZSBJbnB1dAoKPHByZSBjbGFzcz0="brush:java;">2 2 4
Sample Output
6 90
题意就是像蜂巢那样的一个点与六个点相邻，从一个点开始走，走了n步之后回到远点的路线数
用三位数组来暴力DP，dp[i][j][k]代表的是走第i步走到坐标为（j，k）的点的方法数
#include
#include
#include
using namespace std; int dp[22][22][22]={0},xx[6][2]={{1,0},{0,1},{-1,0},{0,-1},{1,1},{-1,-1}}; //因为要回原点，所以最远也只会和原点相距14/2的距离，所以取（7，7）为原点（这个原点是随便去的） void DP() { int i,j,k,l,a,b; dp[0][7][7]=1; //初始化 for(i=1;i<=14;i++) //最多14步 for(j=0;j<=14;j++) //遍历可到的全部坐标点 for(k=0;k<=14;k++) for(l=0;l<6;l++) //遍历相邻的各个方向 dp[i][j][k]+=dp[i-1][j+xx[l][0]][k+xx[l][1]]; } int main (void) { int n,m,i,j,k,l; DP(); scanf("%d",&n); while(n--&&scanf("%d",&m)) { printf("%d\n",dp[m][7][7]); } return 0; }