Double Patience UVA

题目传送门

题意：36张牌分成9堆，每堆4张牌，每一次可以拿走顶部两个相同点数的牌，如果有多种方法则按照等概率的随机拿，如果最后拿走了所有的牌则胜利，求胜利的概率。

思路；用一个九元组储存状态，进行记忆话搜索。

#include <algorithm>#include <cmath>#include <cstdio>#include <cstring>#include <iostream>#include <list>#include <map>#include <queue>#include <set>#include <stack>#include <string>#include <vector>#define MAXN 110#define MAXE 5#define INF 1000000000#define MOD 10007#define LL long long#define ULL unsigned long long#define pi 3.14159using namespace std;double dp[MAXE][MAXE][MAXE][MAXE][MAXE][MAXE][MAXE][MAXE][MAXE];bool vis[MAXE][MAXE][MAXE][MAXE][MAXE][MAXE][MAXE][MAXE][MAXE];vector<char> p[MAXN];double dfs(int num1, int num2, int num3, int num4, int num5, int num6, int num7, int num8, int num9) {    if (vis[num1][num2][num3][num4][num5][num6][num7][num8][num9])        return dp[num1][num2][num3][num4][num5][num6][num7][num8][num9];    vis[num1][num2][num3][num4][num5][num6][num7][num8][num9] = true;    int num[] = {num1, num2, num3, num4, num5, num6, num7, num8, num9};    bool flag = false;    double &ans = dp[num1][num2][num3][num4][num5][num6][num7][num8][num9];    for (int i = 0; i < 9; ++i) {        if (num[i]) {            flag = true;            break;        }    }    if (!flag) {        return ans = 1;    }    int cnt = 0;    for (int i = 0; i < 9; ++i) {        if (!num[i])            continue;        for (int j = i + 1; j < 9; ++j) {            if (!num[j])                continue;            if (p[i][num[i] - 1] == p[j][num[j] - 1]) {                cnt++;                num[i]--;                num[j]--;                ans += dfs(num[0], num[1], num[2], num[3], num[4], num[5], num[6], num[7], num[8]);                num[i]++;                num[j]++;            }        }    }    if (ans == 0)        return ans = 0;    return ans = ans / (double) cnt;}int main() {    std::ios::sync_with_stdio(false);    string str1, str2, str3, str4;    while (cin >> str1 >> str2 >> str3 >> str4) {        memset(dp, 0, sizeof(dp));        memset(vis, false, sizeof(vis));        memset(p, 0, sizeof(p));        p[0].push_back(str1[0]);        p[0].push_back(str2[0]);        p[0].push_back(str3[0]);        p[0].push_back(str4[0]);        for (int i = 1; i <= 8; ++i) {            cin >> str1 >> str2 >> str3 >> str4;            p[i].push_back(str1[0]);            p[i].push_back(str2[0]);            p[i].push_back(str3[0]);            p[i].push_back(str4[0]);        }        printf("%.6f\n", dfs(4, 4, 4, 4, 4, 4, 4, 4, 4));    }    return 0;}/* AS 9S 6C KS JC QH AC KH 7S QD JD KD QS TS JS 9H 6D TD AD 8S QC TH KC 8D 8C 9D TC 7C 9C 7H JH 7D 8H 6S AH 6H */