[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
1 #pragma comment(linker, "/STACK:10240000,10240000") 2 3 #include <iostream> 4 #include <cstdio> 5 #include <algorithm> 6 #include <cstdlib> 7 #include <cstring> 8 #include <map> 9 #include <queue> 10 #include <deque> 11 #include <cmath> 12 #include <vector> 13 #include <ctime> 14 #include <cctype> 15 #include <set> 16 #include <bitset> 17 #include <functional> 18 #include <numeric> 19 #include <stdexcept> 20 #include <utility> 21 22 using namespace std; 23 24 #define mem0(a) memset(a, 0, sizeof(a)) 25 #define mem_1(a) memset(a, -1, sizeof(a)) 26 #define lson l, m, rt << 1 27 #define rson m + 1, r, rt << 1 | 1 28 #define rep_up0(a, b) for (int a = 0; a < (b); a++) 29 #define rep_up1(a, b) for (int a = 1; a <= (b); a++) 30 #define rep_down0(a, b) for (int a = b - 1; a >= 0; a--) 31 #define rep_down1(a, b) for (int a = b; a > 0; a--) 32 #define all(a) (a).begin(), (a).end() 33 #define lowbit(x) ((x) & (-(x))) 34 #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) {} 35 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {} 36 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {} 37 #define pchr(a) putchar(a) 38 #define pstr(a) printf("%s", a) 39 #define sstr(a) scanf("%s", a) 40 #define sint(a) scanf("%d", &a) 41 #define sint2(a, b) scanf("%d%d", &a, &b) 42 #define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c) 43 #define pint(a) printf("%d\n", a) 44 #define test_print1(a) cout << "var1 = " << a << endl 45 #define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl 46 #define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl 47 #define mp(a, b) make_pair(a, b) 48 #define pb(a) push_back(a) 49 50 typedef unsigned int uint; 51 typedef long long LL; 52 typedef pair<int, int> pii; 53 typedef vector<int> vi; 54 55 const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1}; 56 const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 }; 57 const int maxn = 1e3 + 7; 58 const int md = 100000007; 59 const int inf = 1e9 + 7; 60 const LL inf_L = 1e18 + 7; 61 const double pi = acos(-1.0); 62 const double eps = 1e-6; 63 64 template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);} 65 template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;} 66 template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;} 67 template<class T>T condition(bool f, T a, T b){return f?a:b;} 68 template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];} 69 int make_id(int x, int y, int n) { return x * n + y; } 70 71 int dp[1005][2010][2]; 72 73 void init() { 74 dp[1][1][0] = 1; 75 dp[1][2][1] = 1; 76 for (int i = 2; i <= 1000; i ++) { 77 rep_up1(j, 2 * i) { 78 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]; 79 dp[i][j][1] = dp[i - 1][j][1] + dp[i - 1][j - 1][0] * 2 + dp[i - 1][j - 1][1] * 2; 80 if (j > 1) dp[i][j][1] += dp[i - 1][j - 2][0] + dp[i - 1][j - 2][1]; 81 dp[i][j][0] %= md; 82 dp[i][j][1] %= md; 83 } 84 } 85 } 86 87 int main() { 88 //freopen("in.txt", "r", stdin); 89 init(); 90 int T; 91 cin >> T; 92 rep_up0(cas, T) { 93 int n, k; 94 cin >> n >> k; 95 cout << (dp[n][k][0] + dp[n][k][1]) % md << endl; 96 } 97 return 0; 98 }
