hihoCoder 1159 扑克牌 (dp,难)

题意：

一副不含王的扑克牌由52张牌组成，由红桃、黑桃、梅花、方块4组牌组成，每组13张不同的面值。现在给定52张牌中的若干张，请计算将它们排成一列，相邻的牌面值不同的方案数。

牌的表示方法为XY，其中X为面值，为2、3、4、5、6、7、8、9、T、J、Q、K、A中的一个。Y为花色，为S、H、D、C中的一个。如2S、2H、TD等。

题解：

对应每个排有4中类型

dp[a][b][c][d][pre]表示有a个只含1种类型，b个含2两种类型，c个含3种类型，d个含4种类型，前一次选择的是pre种类型的。

比较难想这个状态，转移也并不容易，此题还有其他做法是利用组合数学的方法去dp但并不能懂。

#include<iostream>#include<math.h>#include<stdio.h>#include<algorithm>#include<string.h>#include<string>#include<vector>#include<queue>#include<map>#include<set>#include<stack>#define B(x) (1<<(x))using namespace std;typedef long long ll;typedef unsigned long long ull;const int oo = 0x3f3f3f3f;const ll OO = 0x3f3f3f3f3f3f3f3f;const double eps = 1e-9;#define lson rt<<1#define rson rt<<1|1void cmax(int& a, int b){ if (b > a)a = b; }void cmin(int& a, int b){ if (b < a)a = b; }void cmax(ll& a, ll b){ if (b > a)a = b; }void cmin(ll& a, ll b){ if (b < a)a = b; }void cmax(double& a, double b){ if (a - b < eps) a = b; }void cmin(double& a, double b){ if (b - a < eps) a = b; }void add(int& a, int b, int mod){ a = (a + b) % mod; }void add(ll& a, ll b, ll mod){ a = (a + b) % mod; }const ll MOD = 1000000007;const int maxn = 110000;ull dp[15][15][15][15][15];char pm[60][60];int num[300];ull dfs(int a, int b, int c, int d, int pre){if (dp[a][b][c][d][pre] != -1)return dp[a][b][c][d][pre];ull res = 0;if (a > 0){if (pre != 1) res += a * dfs(a - 1, b, c, d, 0);else res += (a - 1) * dfs(a - 1, b, c, d, 0);}if (b > 0){if (pre != 2) res += 2 * b * dfs(a + 1, b - 1, c, d, 1);else res += 2 * (b - 1) * dfs(a + 1, b - 1, c, d, 1);}if (c > 0){if (pre != 3) res += 3 * c * dfs(a, b + 1, c - 1, d, 2);else res += 3 * (c - 1) * dfs(a, b + 1, c - 1, d, 2);}if (d > 0){if (pre != 4) res += 4 * d * dfs(a, b, c + 1, d - 1, 3);else res += 4 * (d - 1) * dfs(a, b, c + 1, d - 1, 3);}return dp[a][b][c][d][pre] = res;}void Init(){memset(dp, -1, sizeof dp);for (int i = 0; i < 4; i++)dp[0][0][0][0][i] = 1;}int main(){//freopen("E:\\read.txt", "r", stdin);int T, n, a, b, c, d, cas = 1;Init();scanf("%d", &T);while (T--){scanf("%d", &n);memset(num, 0, sizeof num);for (int i = 1; i <= n; i++){scanf("%s", pm[i]);num[pm[i][0]]++;}a = b = c = d = 0;for (int i = 0; i < 300; i++){if (num[i] == 1) a++;if (num[i] == 2) b++;if (num[i] == 3) c++;if (num[i] == 4) d++;}printf("Case #%d: %llu\n", cas++, dfs(a, b, c, d, 0));}return 0;}