結果
問題 | No.309 シャイな人たち (1) |
ユーザー |
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提出日時 | 2018-10-01 09:38:31 |
言語 | C++14 (gcc 13.3.0 + boost 1.87.0) |
結果 |
AC
|
実行時間 | 2,469 ms / 4,000 ms |
コード長 | 2,923 bytes |
コンパイル時間 | 1,873 ms |
コンパイル使用メモリ | 178,860 KB |
実行使用メモリ | 6,820 KB |
最終ジャッジ日時 | 2024-10-12 09:29:01 |
合計ジャッジ時間 | 20,061 ms |
ジャッジサーバーID (参考情報) |
judge3 / judge2 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 13 |
ソースコード
#include <bits/stdc++.h>using namespace std;using VS = vector<string>; using LL = long long;using VI = vector<int>; using VVI = vector<VI>;using PII = pair<int, int>; using PLL = pair<LL, LL>;using VL = vector<LL>; using VVL = vector<VL>;#define ALL(a) begin((a)),end((a))#define RALL(a) (a).rbegin(), (a).rend()#define SZ(a) int((a).size())#define SORT(c) sort(ALL((c)))#define RSORT(c) sort(RALL((c)))#define UNIQ(c) (c).erase(unique(ALL((c))), end((c)))#define FOR(i, s, e) for (int(i) = (s); (i) < (e); (i)++)#define FORR(i, s, e) for (int(i) = (s); (i) > (e); (i)--)#define debug(x) cerr << #x << ": " << x << endlconst int INF = 1e9; const LL LINF = 1e16;const LL MOD = 1000000007; const double PI = acos(-1.0);int DX[8] = { 0, 0, 1, -1, 1, 1, -1, -1 }; int DY[8] = { 1, -1, 0, 0, 1, -1, 1, -1 };/* ----- 2018/09/30 Problem: yukicoder 309 / Link: http://yukicoder.me/problems/no/309 ----- *//* ------問題-----------問題ここまで----- *//* -----解説等---------解説ここまで---- */LL ans = 0LL;int main() {cin.tie(0);ios_base::sync_with_stdio(false);int H, W; cin >> H >> W;vector<vector<double>>P(H, vector<double>(W));FOR(i, 0, H) {FOR(j, 0, W) {cin >> P[i][j];P[i][j] /= 100.0;}}VVI S(H, VI(W));FOR(i, 0, H) {FOR(j, 0, W) {cin >> S[i][j];}}vector<vector<double>>dp(H + 1, vector<double>(1 << W, 0));dp[0][0] = 1;FOR(i, 0, H) {FOR(state, 0, 1 << W) {double p = 1;FOR(j, 0, W) {if (state & 1 << j) {p *= P[i][j];}else {p *= 1 - P[i][j];}}// 現在の列がstateであるような確率pは分かっているauto ok = [](int s, int k) {return s & 1 << k;};FOR(pre, 0, 1 << W) {if (dp[i][pre] < 1e-12)continue;int upstate = 0;FOR(j, 0, W) {if (S[i][j] == 0 && ok(state, j))upstate |= 1 << j;if (S[i][j] == 1 && ok(state, j) && ok(pre, j))upstate |= 1 << j;}FOR(j, 1, W) {if (S[i][j] == 1 && ok(state, j) && ok(upstate, j - 1))upstate |= 1 << j;if (S[i][j] == 2 && ok(state, j) && ok(pre, j) && ok(upstate, j - 1))upstate |= 1 << j;}FORR(j, W - 2, 0 - 1) {if (S[i][j] == 1 && ok(state, j) && ok(upstate, j + 1))upstate |= 1 << j;if (S[i][j] == 2 && ok(state, j) && ok(pre, j) && ok(upstate, j + 1))upstate |= 1 << j;}FORR(j, W - 2, 0) {if (S[i][j] == 2 && ok(state, j) && ok(upstate,j-1) && ok(upstate, j + 1))upstate |= 1 << j;if (S[i][j] == 3 && ok(state, j) && ok(pre, j) && ok(upstate, j - 1) && ok(upstate, j + 1))upstate |= 1 << j;}dp[i + 1][upstate] += p * dp[i][pre];}}}double ans = 0;FOR(i, 0, H) {FOR(state, 0, 1 << W) {ans += __builtin_popcount(state)*dp[i + 1][state];}}cout << fixed << setprecision(10) << ans << "\n";return 0;}