[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
", 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[8] = {0, 0, -1, 1, 1, 1, -1, -1};
const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 };
const int maxn = 1e3 + 7;
const int md = 100000007;
const int inf = 1e9 + 7;
const LL inf_L = 1e18 + 7;
const double pi = acos(-1.0);
const double eps = 1e-6;

template<class T>T gcd(T a, T b){return b==0?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[1005][2010][2];

void init() {
    dp[1][1][0] = 1;
    dp[1][2][1] = 1;
    for (int i = 2; i <= 1000; i ++) {
        rep_up1(j, 2 * i) {
            dp[i][j][0] = dp[i - 1][j][0] + dp[i - 1][j][1] * 2 + dp[i - 1][j - 1][1] + dp[i - 1][j - 1][0];
            dp[i][j][1] = dp[i - 1][j][1] + dp[i - 1][j - 1][0] * 2 + dp[i - 1][j - 1][1] * 2;
            if (j > 1) dp[i][j][1] += dp[i - 1][j - 2][0] + dp[i - 1][j - 2][1];
            dp[i][j][0] %= md;
            dp[i][j][1] %= 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][0] + dp[n][k][1]) % md << endl;
    }
    return 0;
}