結果
問題 | No.309 シャイな人たち (1) |
ユーザー | rpy3cpp |
提出日時 | 2016-02-12 03:11:03 |
言語 | C++11 (gcc 11.4.0) |
結果 |
AC
|
実行時間 | 1,196 ms / 4,000 ms |
コード長 | 4,170 bytes |
コンパイル時間 | 953 ms |
コンパイル使用メモリ | 78,284 KB |
実行使用メモリ | 19,804 KB |
最終ジャッジ日時 | 2024-09-22 04:03:49 |
合計ジャッジ時間 | 15,186 ms |
ジャッジサーバーID (参考情報) |
judge5 / judge1 |
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テストケース
テストケース表示入力 | 結果 | 実行時間 実行使用メモリ |
---|---|---|
testcase_00 | AC | 1,194 ms
19,592 KB |
testcase_01 | AC | 1,194 ms
19,592 KB |
testcase_02 | AC | 1,185 ms
19,592 KB |
testcase_03 | AC | 1,189 ms
19,608 KB |
testcase_04 | AC | 18 ms
6,940 KB |
testcase_05 | AC | 239 ms
7,320 KB |
testcase_06 | AC | 449 ms
19,804 KB |
testcase_07 | AC | 1,190 ms
19,804 KB |
testcase_08 | AC | 1,188 ms
19,588 KB |
testcase_09 | AC | 1,188 ms
19,732 KB |
testcase_10 | AC | 1,193 ms
19,696 KB |
testcase_11 | AC | 1,196 ms
19,732 KB |
testcase_12 | AC | 1,191 ms
19,648 KB |
testcase_13 | AC | 2 ms
6,944 KB |
testcase_14 | AC | 2 ms
6,940 KB |
testcase_15 | AC | 6 ms
7,420 KB |
ソースコード
#include <iostream> #include <vector> #include <deque> #include <utility> #include <iomanip> using namespace std; constexpr double eps = 1e-20; int R = 0; int C = 0; double solve(const int R, const int C, const vector<vector<double>> & P, const vector<vector<int>> & S); vector<int> preprocess(int n_col); int get_next_status(int prev, int know, const vector<int> & s); double calc_prob(int mask, const vector<double> & p); int bitcount(int mask){ int counts = 0; while (mask){ if (mask & 1) ++counts; mask >>= 1; } return counts; } template <typename T> void disp(const vector< vector<T> > & mat){ for (const auto & row : mat){ for (const auto elem : row){ cout << elem << ' '; } cout << endl; } } int main() { cin >> R >> C; vector< vector<int> > S(R, vector<int>(C, 0)); vector< vector<double> > P(R, vector<double>(C, 0.0)); // read data for (auto & row : P){ for (auto & p : row){ cin >> p; p /= 100.0; } } for (auto & row : S){ for (auto & s : row) cin >> s; } // solve cout << fixed << setprecision(12); cout << solve(R, C, P, S) << endl; return 0; } double solve(const int R, const int C, const vector<vector<double>> & P, const vector<vector<int>> & S){ auto next_status = preprocess(C); double ans = 0; int limit = 1 << C; vector<double> old_prob(1 << C, 0.0); vector<double> new_prob(1 << C, 0.0); old_prob[0] = 1.0; for (int r = 0; r != R; ++r){ auto s = S[r]; auto p = P[r]; for (int know = 0; know != limit; ++know){ auto new_p = calc_prob(know, p); if (new_p <= eps) continue; for (int prev = 0; prev != limit; ++prev){ auto new_status = next_status[get_next_status(prev, know, s)]; new_prob[new_status] += old_prob[prev] * new_p; } } for (int status = 0; status != 1 << C; ++status){ ans += new_prob[status] * bitcount(status); } swap(old_prob, new_prob); for (auto & np : new_prob) np = 0; } return ans; } vector<int> preprocess(int n_col){ vector<int> next_status(1 << (2 * n_col), 0); next_status[3] = 1; for (int c = 1; c != n_col; ++c){ int tail_limit = 1 << (2 * c - 2); for (auto head : {1, 2, 3}){ int head_mask = head << (2 * c); int head_result2 = (head >= 2) ? (1 << c) : 0; int head_result_on = 1 << c; for (auto neck : {0, 1, 2, 3}){ int neck_mask = neck << (2 * c - 2); int neck_tail_limit = neck_mask + tail_limit; int neck_threshold = 1 << (c - 1); for (int neck_tail = neck_mask; neck_tail != neck_tail_limit; ++neck_tail){ int prev = next_status[neck_tail]; if (prev >= neck_threshold){ next_status[head_mask | neck_tail] = head_result2 + prev; } else if (head != 3){ next_status[head_mask | neck_tail] = prev; } else { next_status[head_mask | neck_tail] = head_result_on + next_status[neck_tail + tail_limit]; } } } } } return next_status; } double calc_prob(int mask, const vector<double> & p){ double prob = 1.0; for (auto pi : p){ prob *= (mask & 1) ? pi : (1 - pi); mask >>= 1; } return prob; } int get_next_status(int prev, int know, const vector<int> & s){ // prev & (1<<i) : 前列i番目の人が挙手していれば1、挙手していなければ0 // know & (1<<i) : 当列i番目の人が知っていれば1、知らなければ0 // s[i] : 恥ずかしがり度 int ans = 0; for (int i = 0; i != C; ++i){ if ((know & (1 << i)) == 0) continue; int tmp = 3 - s[i]; if (prev & (1 << i)) ++tmp; if (tmp <= 0) continue; if (tmp == 4) tmp = 3; ans += (tmp << (2 * i)); } return ans; }