[hdu4301]DP

题意：给一个2*n的矩形块，求把它分成k个连通块的方法数。（有公共边即视为联通）

思路：由于宽度只有2，于是很容易设计状态使问题满足阶段性以及无后效性。具体来说，令dp[i][j][0]和dp[i][j][1]表示把前i行分成j个连通块最后两个格子分别属于和不属于同一个连通块的方法数，于是有下面的状态转移方程：

dp[i][j][0]=dp[i-1][j-1][0..1]+dp[i-1][j][0]+dp[i-1][j][1]*2

dp[i][j][1]=dp[i-1][j-2][0..1]+dp[i-1][j][1]+dp[i-1][j-1][0..1]*2

 #pragma comment(linker, "/STACK:10240000,10240000")
 
 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <cstdlib>
 #include <cstring>
 #include <map>
 #include <queue>
 #include <deque>
 #include <cmath>
 #include <vector>
 #include <ctime>
 #include <cctype>
 #include <set>
 #include <bitset>
 #include <functional>
 #include <numeric>
 #include <stdexcept>
 #include <utility>
 
 using namespace std;
 
 #define mem0(a) memset(a, 0, sizeof(a))
 #define mem_1(a) memset(a, -1, sizeof(a))
 #define lson l, m, rt << 1
 #define rson m + 1, r, rt << 1 | 1
 #define rep_up0(a, b) for (int a = 0; a < (b); a++)
 #define rep_up1(a, b) for (int a = 1; a <= (b); a++)
 #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--)
 #define rep_down1(a, b) for (int a = b; a > 0; a--)
 #define all(a) (a).begin(), (a).end()
 #define lowbit(x) ((x) & (-(x)))
 #define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {}
 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
 #define pchr(a) putchar(a)
 #define pstr(a) printf("%s", a)
 #define sstr(a) scanf("%s", a)
 #define sint(a) scanf("%d", &a)
 #define sint2(a, b) scanf("%d%d", &a, &b)
 #define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c)
 #define pint(a) printf("%d\n", a)
 #define test_print1(a) cout << "var1 = " << a << endl
 #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl
 #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl
 #define mp(a, b) make_pair(a, b)
 #define pb(a) push_back(a)
 
 typedef unsigned int uint;
 typedef long long LL;
 typedef pair<int, int> pii;
 typedef vector<int> vi;
 
 const int dx[] = {, , -, , , , -, -};
 const int dy[] = {-, , , , , -, , - };
 const int maxn = 1e3 + ;
 const int md = ;
 const int inf = 1e9 + ;
 const LL inf_L = 1e18 + ;
 const double pi = acos(-1.0);
 const double eps = 1e-;
 
 template<class T>T gcd(T a, T b){return b==?a:gcd(b,a%b);}
 template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;}
 template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;}
 template<class T>T condition(bool f, T a, T b){return f?a:b;}
 template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];}
 int make_id(int x, int y, int n) { return x * n + y; }
 
 int dp[][][];
 
 void init() {
     dp[][][] = ;
     dp[][][] = ;
     for (int i = ; i <= ; i ++) {
         rep_up1(j,  * i) {
             dp[i][j][] = dp[i - ][j][] + dp[i - ][j][] *  + dp[i - ][j - ][] + dp[i - ][j - ][];
             dp[i][j][] = dp[i - ][j][] + dp[i - ][j - ][] *  + dp[i - ][j - ][] * ;
             if (j > ) dp[i][j][] += dp[i - ][j - ][] + dp[i - ][j - ][];
             dp[i][j][] %= md;
             dp[i][j][] %= md;
         }
     }
 }
 
 int main() {
     //freopen("in.txt", "r", stdin);
     init();
     int T;
     cin >> T;
     rep_up0(cas, T) {
         int n, k;
         cin >> n >> k;
         cout << (dp[n][k][] + dp[n][k][]) % md << endl;
     }
     return ;
 }