最近常登《美食大战老鼠》这款沙雕游戏,不得不赞叹道里面的卡片强化系统真的****。为了最大化用有限卡裸上出高星卡的成功概率,我就瞎写了这么个大暴力程序。
算法上,先用了个不能再沙雕的裸搜索(用哈希存状态),然后上个无聊的大 dp 就完事儿了。
用下面的代码实际操作了几次后感觉最优策略似乎有规律可循,所以应该还有提速空间。代码本身也有一些可优化的地方,不过懒得改了。
基本使用方式:先输入 \(n\) 表示你已有的卡片数量,再输入 \(n\) 个数表示 \(n\) 张卡片的星级(顺序可任意),最后输入一个数表示你希望强化出的卡片星级。每一步程序都会输出最终成功概率以及当前的最优决策,按要求执行反馈即可。
注意事项:
- 此代码未对任何非法操作做判断,所以请尽量保证输入合法(比如,不要出现希望强化出的卡片星级低于已有卡片的情况)
- 由于算法过于暴力,所以卡片数量不宜太多(最好不超过 \(50\) 张)
- 下面的概率表默认强化用卡为高耗能卡(你应该明白是什么意思)
- 强化默认所使用的是同一种四叶草(对应的程序语句为
const double mul = 1.0;
,mul
表示四叶草对应的概率倍数,例如:1.0
为不使用四叶草,使用 1 级四叶草则将1.0
改为1.2
,以此类推) - 由于未找到 15 星及以上强化对应的概率,因此下面的代码只适用于 15 级以下的强化
(我并不知道强化的基本概率的计算公式,所以只能打表) - 再次强调,是裸上(即用单卡强化
,因为我平时就单上,懒得写复杂情况了) 可能有bug
#include<bits/stdc++.h>
using namespace std;
const int N = 1000000;
const int M = 16;
const int mod = 998244353;
const double mul = 1.0;
const double eps = 1e-8;
int n, all, goal;
double p[M][M], f[N];
bool visit[N];
map<int, int> hash_t;
pair<int, int> best[N];
void init() {
p[0][0] = 1;
p[1][1] = 1;
p[2][2] = 0.9683;
p[3][3] = 0.6858;
p[4][4] = 0.495;
p[5][5] = 0.3958;
p[6][6] = 0.3192;
p[7][7] = 0.2642;
p[8][8] = 0.22;
p[9][9] = 0.135;
p[10][10] = 0.125;
p[11][11] = 0.116;
p[12][12] = 0.107;
p[13][13] = 0.101;
p[14][14] = 0.095;
p[1][0] = 0.88;
p[2][1] = 0.792;
p[3][2] = 0.55;
p[4][3] = 0.4033;
p[5][4] = 0.33;
p[6][5] = 0.264;
p[7][6] = 0.212;
p[8][7] = 0.132;
p[9][8] = 0.045;
p[10][9] = 0.044;
p[11][10] = 0.043;
p[12][11] = 0.0398;
p[13][12] = 0.0367;
p[14][13] = 0.0336;
p[2][0] = 0.6083;
p[3][1] = 0.4292;
p[4][2] = 0.2417;
p[5][3] = 0.2008;
p[6][4] = 0.132;
p[7][5] = 0.106;
p[8][6] = 0.06;
p[9][7] = 0.022;
p[10][8] = 0.018;
p[11][9] = 0.017;
p[12][10] = 0.0156;
p[13][11] = 0.0141;
p[14][12] = 0.0126;
for (int i = 0; i < 15; ++i) {
p[i][i] *= mul;
p[i][i] = min(p[i][i], 1.);
}
for (int i = 1; i < 15; ++i) {
p[i][i - 1] *= mul;
p[i][i - 1] = min(p[i][i - 1], 1.);
}
for (int i = 2; i < 15; ++i) {
p[i][i - 2] *= mul;
p[i][i - 2] = min(p[i][i - 2], 1.);
}
}
pair<bool, int> decode(vector<int> info) {
bool exist = true;
int base = 0;
for (auto x : info) {
base = (base * 233ll + x) % mod;
}
if (!hash_t.count(base)) {
exist = false;
hash_t[base] = ++all;
if (info[goal]) {
visit[all] = true;
f[all] = 1;
}
}
return make_pair(exist, hash_t[base]);
}
void strengthen(int i, int j, vector<int>& number, bool success = true) {
--number[i];
--number[j];
if (success) {
++number[i + 1];
} else {
++number[i - (i > 5)];
}
}
int count(vector<int> number) {
int result = 0;
for (auto x : number) {
result += x;
}
return result;
}
void dfs(vector<int> number) {
bool exist = decode(number).first;
if (count(number) != 1 && exist == false) {
for (int i = 0; i < 15; ++i) {
for (int j = max(0, i - 2); j < i; ++j) {
if (number[i] && number[j]) {
vector<int> next_array = number;
strengthen(i, j, next_array);
dfs(next_array);
next_array = number;
strengthen(i, j, next_array, false);
dfs(next_array);
}
}
if (number[i] >= 2) {
vector<int> next_array = number;
strengthen(i, i, next_array);
dfs(next_array);
next_array = number;
strengthen(i, i, next_array, false);
dfs(next_array);
}
}
}
}
double dp(vector<int> number) {
int id = decode(number).second;
if (visit[id]) {
return f[id];
}
visit[id] = true;
if (count(number) == 1) {
return f[id] = 0;
} else {
double& answer = f[id];
for (int i = 0; i < 15; ++i) {
for (int j = max(0, i - 2); j < i; ++j) {
if (number[i] && number[j]) {
vector<int> array1 = number, array0 = number;
strengthen(i, j, array1);
strengthen(i, j, array0, false);
double foo = dp(array1) * p[i][j] + dp(array0) * (1 - p[i][j]);
if (foo > answer) {
best[id] = make_pair(i, j);
answer = foo;
}
}
}
if (number[i] >= 2) {
vector<int> array1 = number, array0 = number;
strengthen(i, i, array1);
strengthen(i, i, array0, false);
double foo = dp(array1) * p[i][i] + dp(array0) * (1 - p[i][i]);
if (foo > answer) {
best[id] = make_pair(i, i);
answer = foo;
}
}
}
return answer;
}
}
int main() {
init();
cin >> n;
vector<int> number(16);
for (int i = 1; i <= n; ++i) {
int x;
cin >> x;
++number[x];
}
cin >> goal;
dfs(number);
//freopen("log.txt", "w", stdout);
assert(all < N);
//cerr << all << '\n';
cout << "probability: " << setprecision(10) << dp(number) << '\n';
if (dp(number) > eps) {
while (1) {
int id = decode(number).second;
cout << "the best choice is: strengthen " << best[id].first << " with " << best[id].second << '.' << '\n';
cout << "have you succeeded? (input 1 for success and 0 for failure)" << '\n';
string inf;
cin >> inf;
strengthen(best[id].first, best[id].second, number, inf == "1");
id = decode(number).second;
if (f[id] == 1) {
cout << "congratulation!" << '\n';
break;
} else if (f[id] < eps) {
cout << "sorry, it's impossible now." << '\n';
break;
} else {
cout << "probability now: " << setprecision(10) << f[id] << '\n';
}
}
} else {
cout << "it's impossible." << '\n';
}
cout << "done." << '\n';
return 0;
}