[不平等博弈][DP] TCO 2017 Final Div1 Medium GameOfTokens-优快云博客

本文链接：https://blog.youkuaiyun.com/Vectorxj/article/details/79350156

本文介绍了一种针对游戏令牌的计数算法，该算法通过动态规划解决令牌模式字符串中的A和B数量变化问题，旨在计算A比B多走的步数，并给出具体的实现代码和测试案例。

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$Solution$

先考虑已知模板串的情况。
如果出现ABB这样A的个数比B的个数少的，那么A是不可能会向左走的。
整个模板串就会因为这个分隔开成若干个独立游戏。
我们要计算每个游戏的值的和。
维护两个变量 $ans$ 和 $cur$ 分别表示A能比B多走 $ans$ 步，当前游戏A的个数比B的个数多几个。
显然的有以下三个转移：

$pattern_i=$ A， $cur = max(cur + 1, 0)$ 。
$pattern_i=$ B， $cur = min(cur - 1, 0)$ 。
$pattern_i=$ .， $ans = ans + cur$ 。

求方案数就很简单了 $dp_{i,ans,cur}$ 把所有状态记下来就好了。
~~数组开小sb了qwq~~

// BEGIN CUT HERE

// END CUT HERE
#line 5 "GameOfTokens.cpp"
#include <bits/stdc++.h>
using namespace std;

const int N = 52;
const int M = 5050;
const int MOD = 1000000007;

int dp[N][M][2 * N];
char p[N];
int n, m;

class GameOfTokens {  
public:
    inline void Add(int &x, int a) {
        x = (x + a) % MOD;
    }
    inline int &f(int i, int j, int k) {
        return dp[i][j + 2501][k + 51];
    }
    int count(string pattern) {
        memset(dp, 0, sizeof dp);
        n = pattern.size(); m = n * n;
        for (int i = 1; i <= n; i++)
            p[i] = pattern[i - 1];
        if (p[n] == 'A' || p[n] == '?') f(n, 0, 1) = 1;
        if (p[n] == 'B' || p[n] == '?') f(n, 0, -1) = 1;
        if (p[n] == '.' || p[n] == '?') f(n, 0, 0) = 1;
        for (int i = n - 1; ~i; i--) {
            for (int j = -m; j <= m; j++)
                for (int k = -n; k <= n; k++) {
                    int res = f(i + 1, j, k);
                    if (!res) continue;
                    if (p[i] == 'A' || p[i] == '?') {
                        int ans = j;
                        int cur = max(k + 1, 0);
                        Add(f(i, ans, cur), res);
                    }
                    if (p[i] == 'B' || p[i] == '?') {
                        int ans = j;
                        int cur = min(k - 1, 0);
                        Add(f(i, ans, cur), res);
                    }
                    if (p[i] == '.' || p[i] == '?') {
                        int cur = k;
                        int ans = j + cur;
                        Add(f(i, ans, cur), res);
                    }
                }
        }
        int ans = 0;
        for (int j = 1; j <= m; j++)
            for (int k = -n; k <= n; k++)
                Add(ans, f(1, j, k));
        return ans;
    }

    // BEGIN CUT HERE
public:
    void run_test(int Case) { if ((Case == -1) || (Case == 0)) test_case_0(); if ((Case == -1) || (Case == 1)) test_case_1(); if ((Case == -1) || (Case == 2)) test_case_2(); if ((Case == -1) || (Case == 3)) test_case_3(); if ((Case == -1) || (Case == 4)) test_case_4(); }
private:
    template <typename T> string print_array(const vector<T> &V) { ostringstream os; os << "{ "; for (typename vector<T>::const_iterator iter = V.begin(); iter != V.end(); ++iter) os << '\"' << *iter << "\","; os << " }"; return os.str(); }
    void verify_case(int Case, const int &Expected, const int &Received) { cerr << "Test Case #" << Case << "..."; if (Expected == Received) cerr << "PASSED" << endl; else { cerr << "FAILED" << endl; cerr << "\tExpected: \"" << Expected << '\"' << endl; cerr << "\tReceived: \"" << Received << '\"' << endl; } }
    void test_case_0() { string Arg0 = ".AB"; int Arg1 = 0; verify_case(0, Arg1, count(Arg0)); }
    void test_case_1() { string Arg0 = "???B"; int Arg1 = 1; verify_case(1, Arg1, count(Arg0)); }
    void test_case_2() { string Arg0 = "???????...BBB"; int Arg1 = 0; verify_case(2, Arg1, count(Arg0)); }
    void test_case_3() { string Arg0 = "?.A?.B??.??B?"; int Arg1 = 517; verify_case(3, Arg1, count(Arg0)); }
    void test_case_4() { string Arg0 = "????????????????????"; int Arg1 = 612621096; verify_case(4, Arg1, count(Arg0)); }

    // END CUT HERE

};

// BEGIN CUT HERE
int main(void) {
    GameOfTokens ___test;
    ___test.run_test(-1);
    system("pause");
}
// END CUT HERE