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// Source: http://codeforces.com/blog/entry/13225 // Non-strict. int lis_non_strict(const vector<int>& a) { multiset<int> s; for (int x : a) { s.insert(x); auto it = s.upper_bound(x); if (it != s.end()) s.erase(it); } return s.size(); } // Strict. int lis_strict(const vector<int>& a) { multiset<int> s; for (int x : a) { s.insert(x); auto it = s.lower_bound(x); it++; if (it != s.end()) s.erase(it); } return s.size(); } // Return indices of LIS (strict) vector<int> lis_strict_trace(const vector<int>& a) { int n = (int) a.size(); vector<int> b(n+1, 0), f(n, 0); int answer = 0; for (int i = 0; i < n; i++) { f[i] = lower_bound(b.begin() + 1, b.begin()+answer+1, a[i]) - b.begin(); answer = max(answer, f[i]); b[f[i]] = a[i]; } int require = answer; vector<int> T; for (int i = n-1; i >= 0; i--) { if (f[i] == require) { T.push_back(i); require--; } } reverse(T.begin(), T.end()); return T; } // Count number of LIS using mint = long long; // Cnt is exponential. Check if statement says ModInt here? // Returns: (length of LIS, number of LIS) pair<int,mint> count_lis(const vector<int>& a) { if (a.empty()) { return {0, 1}; } // dp[i] = [ (last value, accumulate count) ] for increasing seq of // length i+1 // last value are decreasing vector<vector<pair<int,mint>>> dp(a.size() + 1); int max_len = 0; // returns true if we can append `val` to LIS stored at `cur`. auto pred_len = [] (const vector<pair<int, mint>>& cur, int val) { return !cur.empty() && cur.back().first < val; }; // returns true if we can append `val` after the LIS represented with `p`. auto pred_val = [] (int val, const pair<int,mint>& p) { return val > p.first; }; for (int x : a) { int len = lower_bound(dp.begin(), dp.end(), x, pred_len) - dp.begin(); mint cnt = 1; if (len >= 1) { int pos = upper_bound(dp[len-1].begin(), dp[len-1].end(), x, pred_val) - dp[len-1].begin(); cnt = dp[len-1].back().second; cnt -= (pos == 0) ? 0 : dp[len-1][pos-1].second; } dp[len].emplace_back(x, cnt + (dp[len].empty() ? 0 : dp[len].back().second)); max_len = max(max_len, len + 1); } assert(max_len > 0); return { max_len, dp[max_len-1].back().second, }; }
#line 1 "DP/lis.h" // Source: http://codeforces.com/blog/entry/13225 // Non-strict. int lis_non_strict(const vector<int>& a) { multiset<int> s; for (int x : a) { s.insert(x); auto it = s.upper_bound(x); if (it != s.end()) s.erase(it); } return s.size(); } // Strict. int lis_strict(const vector<int>& a) { multiset<int> s; for (int x : a) { s.insert(x); auto it = s.lower_bound(x); it++; if (it != s.end()) s.erase(it); } return s.size(); } // Return indices of LIS (strict) vector<int> lis_strict_trace(const vector<int>& a) { int n = (int) a.size(); vector<int> b(n+1, 0), f(n, 0); int answer = 0; for (int i = 0; i < n; i++) { f[i] = lower_bound(b.begin() + 1, b.begin()+answer+1, a[i]) - b.begin(); answer = max(answer, f[i]); b[f[i]] = a[i]; } int require = answer; vector<int> T; for (int i = n-1; i >= 0; i--) { if (f[i] == require) { T.push_back(i); require--; } } reverse(T.begin(), T.end()); return T; } // Count number of LIS using mint = long long; // Cnt is exponential. Check if statement says ModInt here? // Returns: (length of LIS, number of LIS) pair<int,mint> count_lis(const vector<int>& a) { if (a.empty()) { return {0, 1}; } // dp[i] = [ (last value, accumulate count) ] for increasing seq of // length i+1 // last value are decreasing vector<vector<pair<int,mint>>> dp(a.size() + 1); int max_len = 0; // returns true if we can append `val` to LIS stored at `cur`. auto pred_len = [] (const vector<pair<int, mint>>& cur, int val) { return !cur.empty() && cur.back().first < val; }; // returns true if we can append `val` after the LIS represented with `p`. auto pred_val = [] (int val, const pair<int,mint>& p) { return val > p.first; }; for (int x : a) { int len = lower_bound(dp.begin(), dp.end(), x, pred_len) - dp.begin(); mint cnt = 1; if (len >= 1) { int pos = upper_bound(dp[len-1].begin(), dp[len-1].end(), x, pred_val) - dp[len-1].begin(); cnt = dp[len-1].back().second; cnt -= (pos == 0) ? 0 : dp[len-1][pos-1].second; } dp[len].emplace_back(x, cnt + (dp[len].empty() ? 0 : dp[len].back().second)); max_len = max(max_len, len + 1); } assert(max_len > 0); return { max_len, dp[max_len-1].back().second, }; }