結果
問題 | No.808 Kaiten Sushi? |
ユーザー |
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提出日時 | 2020-12-08 14:19:26 |
言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
結果 |
TLE
|
実行時間 | - |
コード長 | 5,037 bytes |
コンパイル時間 | 2,771 ms |
コンパイル使用メモリ | 210,636 KB |
最終ジャッジ日時 | 2025-01-16 19:41:05 |
ジャッジサーバーID (参考情報) |
judge4 / judge2 |
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ファイルパターン | 結果 |
---|---|
sample | AC * 3 |
other | AC * 33 TLE * 23 |
コンパイルメッセージ
main.cpp: In function ‘int main()’: main.cpp:172:22: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result] 172 | scanf("%lld",&V[i].first); | ~~~~~^~~~~~~~~~~~~~~~~~~~ main.cpp:177:22: warning: ignoring return value of ‘int scanf(const char*, ...)’ declared with attribute ‘warn_unused_result’ [-Wunused-result] 177 | scanf("%lld",&V[i+N].first); | ~~~~~^~~~~~~~~~~~~~~~~~~~~~
ソースコード
#include <bits/stdc++.h> #include <algorithm> #ifdef _MSC_VER #include <intrin.h> #endif namespace atcoder { namespace internal { // @param n `0 <= n` // @return minimum non-negative `x` s.t. `n <= 2**x` int ceil_pow2(int n) { int x = 0; while ((1U << x) < (unsigned int)(n)) x++; return x; } // @param n `1 <= n` // @return minimum non-negative `x` s.t. `(n & (1 << x)) != 0` int bsf(unsigned int n) { #ifdef _MSC_VER unsigned long index; _BitScanForward(&index, n); return index; #else return __builtin_ctz(n); #endif } } // namespace internal } // namespace atcoder #include <cassert> #include <vector> namespace atcoder { template <class S, S (*op)(S, S), S (*e)()> struct segtree { public: segtree() : segtree(0) {} segtree(int n) : segtree(std::vector<S>(n, e())) {} segtree(const std::vector<S>& v) : _n(int(v.size())) { log = internal::ceil_pow2(_n); size = 1 << log; d = std::vector<S>(2 * size, e()); for (int i = 0; i < _n; i++) d[size + i] = v[i]; for (int i = size - 1; i >= 1; i--) { update(i); } } void set(int p, S x) { assert(0 <= p && p < _n); p += size; d[p] = x; for (int i = 1; i <= log; i++) update(p >> i); } S get(int p) { assert(0 <= p && p < _n); return d[p + size]; } S prod(int l, int r) { assert(0 <= l && l <= r && r <= _n); S sml = e(), smr = e(); l += size; r += size; while (l < r) { if (l & 1) sml = op(sml, d[l++]); if (r & 1) smr = op(d[--r], smr); l >>= 1; r >>= 1; } return op(sml, smr); } S all_prod() { return d[1]; } template <bool (*f)(S)> int max_right(int l) { return max_right(l, [](S x) { return f(x); }); } template <class F> int max_right(int l, F f) { assert(0 <= l && l <= _n); assert(f(e())); if (l == _n) return _n; l += size; S sm = e(); do { while (l % 2 == 0) l >>= 1; if (!f(op(sm, d[l]))) { while (l < size) { l = (2 * l); if (f(op(sm, d[l]))) { sm = op(sm, d[l]); l++; } } return l - size; } sm = op(sm, d[l]); l++; } while ((l & -l) != l); return _n; } template <bool (*f)(S)> int min_left(int r) { return min_left(r, [](S x) { return f(x); }); } template <class F> int min_left(int r, F f) { assert(0 <= r && r <= _n); assert(f(e())); if (r == 0) return 0; r += size; S sm = e(); do { r--; while (r > 1 && (r % 2)) r >>= 1; if (!f(op(d[r], sm))) { while (r < size) { r = (2 * r + 1); if (f(op(d[r], sm))) { sm = op(d[r], sm); r--; } } return r + 1 - size; } sm = op(d[r], sm); } while ((r & -r) != r); return 0; } private: int _n, size, log; std::vector<S> d; void update(int k) { d[k] = op(d[2 * k], d[2 * k + 1]); } }; } // namespace atcoder using namespace atcoder; using namespace std; #define rep(i,n) for (int i = 0; i < (n); ++i) #define Inf 1000000000000000000 vector<int> op(vector<int> a,vector<int> b){ rep(i,2)a[i] = max(a[i],b[i]); return a; } vector<int> e(){ return vector<int>(2,-1); } int main(){ int N; cin>>N; long long L; cin>>L; vector<pair<long long,int>> V(2*N); rep(i,N){ scanf("%lld",&V[i].first); V[i].second = 0; } rep(i,N){ scanf("%lld",&V[i+N].first); V[i+N].second = 1; } sort(V.begin(),V.end()); segtree<vector<int>,op,e> seg(4*N); rep(i,2*N){ vector<int> temp(2,-1); temp[V[i].second] = i; seg.set(i,temp); temp[V[i].second] = 2*N+i; seg.set(2*N+i,temp); } vector<int> ans; int now = 0; while(ans.size()!=2*N){ int temp = 0; if(ans.size()>0)temp = ans.back(); int ok = 4*N,ng = temp; while(ok-ng>1){ int mid = (ok+ng)/2; vector<int> ret = seg.prod(temp,mid); if(ret[now]==-1)ng = mid; else ok = mid; } int x; { vector<int> ret = seg.prod(temp,ok); x = ret[now^1]; } ok = temp,ng = 4*N; while(ng-ok>1){ int mid = (ok+ng)/2; vector<int> ret = seg.prod(temp,mid); if(ret[now^1]<=x)ok = mid; else ng = mid; } vector<int> ret = seg.prod(temp,ok); x = ret[now]%(2*N); ans.push_back(x); seg.set(x,vector<int>(2,-1)); seg.set(x+2*N,vector<int>(2,-1)); now ^= 1; } long long Ans = 0LL; rep(i,2*N){ if(i==0)Ans += V[ans[i]].first; else{ long long temp = V[ans[i]].first - V[ans[i-1]].first; if(temp<0)temp += L; Ans += temp; } } cout<<Ans<<endl; return 0; }