LeetCode935

class Solution935
{
public:

	map<int, vector<int> > gototables = {}; //从一个键可以到达的键的表
	long long Mod = 1000000007;
	int knightDialer(int N)
	{
		init();
		if (N == 1)
		{
			return 10;
		}			
		long long ret=0;
		vector<long long> nums; //记录拨号第M次后各个数字的个数
		for (int i = 0; i < 10; i++)
		{
			nums.emplace_back(0);
		}
		for (int i = 1; i < N; i++)
		{
			if (i == 1)
			{
				for (int j = 0; j < nums.size(); j++)  //第一次拨号后的每个数字个数
				{
					nums[j] += this->gototables[j].size();
				}
			}
			else
			{				
				vector<long long> pre;  //记录前一次的数字个数，用来更新本次拨号后的未位各个数字个数
				for (int t = 0; t < nums.size(); t++)
				{
					pre.emplace_back(nums[t]);
				}
				for (int j = 0; j < nums.size(); j++)
				{
					vector<int> tmp = this->gototables[j];  //该位数字可以按到的下一个，同时也是可以按到该位数字的前一个数字
					long long local = 0;
					for (int m = 0; m < tmp.size(); m++)
					{
						local += pre[tmp[m]];  //正好用0-9下标存储对应数字的个数，直接累加即可
					}
					local %= this->Mod; //余数定理可以先求余后累加
					nums[j] = local; //得到该位数字的个数
				}			
			}
		}
		ret = 0;
		for (int i = 0; i < nums.size(); i++)
		{
			ret += nums[i];  //各个数字为最终尾数的个数和即为结果
		}
		ret %= this->Mod;
		return ret;
	}

	void init()
	{
		gototables[0] = vector < int > {4, 6};
		gototables[1] = vector < int > {8, 6};
		gototables[2] = vector < int > {7, 9};
		gototables[3] = vector < int > {4, 8};
		gototables[4] = vector < int > {3, 9, 0};
		gototables[5] = vector < int > {};
		gototables[6] = vector < int > {1, 7, 0};
		gototables[7] = vector < int > {2, 6};
		gototables[8] = vector < int > {1, 3};
		gototables[9] = vector < int > {2, 4};
	}
};　　

typedef long long ll;
typedef vector<int> VI;
typedef pair<int,int> PII;

#define REP(i,s,t) for(int i=(s);i<(t);i++)
#define FILL(x,v) memset(x,v,sizeof(x))

const int INF = (int)1E9;
#define MAXN 100005

VI adj[10];
const int mod = 1000000007;
int dp[5005][10];
class Solution {
public:
  int knightDialer(int N) {
    adj[1] = VI({6,8});
    adj[2] = VI({7,9});
    adj[3] = VI({4,8});
    adj[4] = VI({0,3,9});
    adj[6] = VI({0,1,7});
    adj[7] = VI({2,6});
    adj[8] = VI({1,3});
    adj[9] = VI({2,4});
    adj[0] = VI({4,6});
    FILL(dp, 0);
    REP(d,0,10) dp[1][d] = 1;
    REP(i,2,N+1) {
      REP(d,0,10) {
        int pre = dp[i-1][d];
        if (pre == 0) continue;
        REP(k,0,adj[d].size()) {
          int d2 = adj[d][k];
          dp[i][d2] = (dp[i][d2] + pre) % mod;
        }
      }
    }
    int ans = 0;
    REP(d,0,10) ans = (ans + dp[N][d]) % mod;
    return ans;
  }
};