結果
| 問題 |
No.5022 XOR Printer
|
| コンテスト | |
| ユーザー |
tnktsyk
|
| 提出日時 | 2025-07-26 15:36:46 |
| 言語 | C++17 (gcc 13.3.0 + boost 1.87.0) |
| 結果 |
AC
|
| 実行時間 | 465 ms / 2,000 ms |
| コード長 | 7,084 bytes |
| コンパイル時間 | 2,387 ms |
| コンパイル使用メモリ | 213,056 KB |
| 実行使用メモリ | 7,716 KB |
| スコア | 4,648,420,920 |
| 最終ジャッジ日時 | 2025-07-26 15:37:14 |
| 合計ジャッジ時間 | 27,212 ms |
|
ジャッジサーバーID (参考情報) |
judge5 / judge6 |
| 純コード判定しない問題か言語 |
(要ログイン)
| ファイルパターン | 結果 |
|---|---|
| other | AC * 50 |
ソースコード
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
/*************************************************
* XOR Printer (yukicoder No.5022)
* Multi-stage greedy with strict non-degradation:
* 1. Compute best single-s solution (2^20 exhaustive), build its ops (baseline).
* 2. Try repeating stages:
* - On current board, find best s by 2^20 exhaustive.
* - Build diff with XOR basis, generate C list.
* - Choose W cells (those gaining with new s).
* - If gain>0 AND total ops within T, commit; else stop.
* 3. Compare final multi-stage score vs baseline. Output better.
*
* Moves only over necessary cells (C/W) with greedy routing to save ops.
* Always safe: never outputs worse score than baseline.
*************************************************/
// ===== XOR Basis (20-bit) with bitset representation =====
struct XorBasis {
static const int B = 20;
int v[B];
bitset<128> used[B];
XorBasis(){ memset(v,0,sizeof(v)); }
void add(int x, const bitset<128>& bs){
bitset<128> c = bs;
for(int b=B-1;b>=0;--b){
if(((x>>b)&1)==0) continue;
if(v[b]==0){ v[b]=x; used[b]=c; return; }
x ^= v[b];
c ^= used[b];
}
}
pair<int, bitset<128>> build(int target) const{
int t=target, made=0; bitset<128> sel; sel.reset();
for(int b=B-1;b>=0;--b){
if(((t>>b)&1)==0) continue;
if(v[b]==0) continue;
t ^= v[b]; made ^= v[b]; sel ^= used[b];
}
return {made, sel};
}
};
inline ll eval_single(const vector<int>& A, int s){
ll sum=0; for(int x:A){ int y=x^s; sum += (y>x?y:x); } return sum; }
// Greedy visit order for a set of indices; append opch for each cell visited.
static void visit_cells(int N, int &r, int &c, const vector<int>& ids, vector<char>& ops, char opch){
if(ids.empty()) return;
vector<char> used(ids.size(),0);
for(size_t done=0; done<ids.size(); ++done){
int best=-1, bestd=1e9;
for(size_t i=0;i<ids.size();++i){
if(used[i]) continue;
int id=ids[i]; int tr=id/N, tc=id%N;
int d=abs(tr-r)+abs(tc-c);
if(d<bestd){ bestd=d; best=i; }
}
used[best]=1;
int id=ids[best]; int tr=id/N, tc=id%N;
while(r<tr){ ops.push_back('D'); ++r; }
while(r>tr){ ops.push_back('U'); --r; }
while(c<tc){ ops.push_back('R'); ++c; }
while(c>tc){ ops.push_back('L'); --c; }
ops.push_back(opch);
}
}
// Convert bitset<128> to vector<int> (0..NN-1)
static vector<int> bs_to_vec(const bitset<128>& bs, int NN){
vector<int> v; v.reserve(NN);
for(int i=0;i<NN;i++) if(bs.test(i)) v.push_back(i);
return v;
}
int main(){
ios::sync_with_stdio(false); cin.tie(nullptr);
int N,T; if(!(cin>>N>>T)) return 0; // N=10, T=1000
const int NN = N*N;
vector<vector<int>> A2(N, vector<int>(N));
for(int i=0;i<N;i++) for(int j=0;j<N;j++) cin>>A2[i][j];
vector<int> flatA(NN);
for(int i=0;i<N;i++) for(int j=0;j<N;j++) flatA[i*N+j]=A2[i][j];
// ---------- Baseline single-s ----------
int bestS_single = 0; ll bestScore_single = -1;
for(int s=0; s<(1<<20); ++s){
ll sc = eval_single(flatA, s);
if(sc > bestScore_single){ bestScore_single=sc; bestS_single=s; }
}
// Build ops for baseline later (after we know if multi-stage beats it)
auto build_baseline_ops = [&](int S)->vector<char>{
XorBasis B; // basis on original board
for(int i=0;i<N;i++) for(int j=0;j<N;j++){
bitset<128> bs; bs.reset(); bs.set(i*N+j);
B.add(A2[i][j], bs);
}
auto [S_real, cells] = B.build(S);
if(S_real != S){
ll sc_real = eval_single(flatA, S_real);
if(sc_real >= bestScore_single){ bestScore_single=sc_real; S=S_real; }
else { auto tmp=B.build(S); cells = tmp.second; }
}
vector<int> Cvec = bs_to_vec(cells, NN);
vector<int> Wvec; Wvec.reserve(NN);
for(int id=0; id<NN; ++id){ int a=flatA[id], b=a^S; if(b>a) Wvec.push_back(id); }
vector<char> ops; ops.reserve(1200);
int r=0,c=0;
visit_cells(N,r,c,Cvec,ops,'C');
visit_cells(N,r,c,Wvec,ops,'W');
if((int)ops.size()>T) ops.resize(T);
return ops;
};
// ---------- Multi-stage attempt ----------
vector<int> board = flatA; // current values
int curS = 0; // current s
vector<char> ops_multi; // operations for multi-stage plan
int r=0,c=0; // cursor pos
ll currentScore = 0; for(int v:board) currentScore += v;
auto time_start = chrono::steady_clock::now();
const double TIME_LIMIT = 1.90; // safety
while(true){
if((int)ops_multi.size() > T-200) break; // margin
double t = chrono::duration<double>(chrono::steady_clock::now()-time_start).count();
if(t > TIME_LIMIT) break;
// 1) best s on current board (exhaustive)
int targetS=-1; ll bestGain=0; // gain over currentScore
for(int s=0; s<(1<<20); ++s){
ll sum=0; for(int x:board){ int y=x^s; sum += (y>x?y:x); }
ll g = sum - currentScore;
if(g > bestGain){ bestGain=g; targetS=s; }
}
if(bestGain <= 0) break; // done
// 2) Build basis on current board, realize diff
int diff = curS ^ targetS;
XorBasis B;
for(int i=0;i<N;i++) for(int j=0;j<N;j++){
bitset<128> bs; bs.reset(); bs.set(i*N+j);
B.add(board[i*N+j], bs);
}
auto [diffReal, needCbs] = B.build(diff);
int newS = curS ^ diffReal;
// 3) choose W cells and compute new score with realizable newS
vector<int> Wvec; Wvec.reserve(NN);
ll newScore = 0;
for(int id=0; id<NN; ++id){
int a = board[id]; int b = a ^ newS;
if(b > a){ Wvec.push_back(id); newScore += b; } else newScore += a;
}
if(newScore <= currentScore) break; // cannot improve actually
// Build ops for this stage to check T first
vector<char> tmp; tmp.reserve(needCbs.count()*6 + Wvec.size()*6 + 10);
int tr=r, tc=c;
vector<int> Cvec = bs_to_vec(needCbs, NN);
visit_cells(N, tr, tc, Cvec, tmp, 'C');
curS = newS; // after C, we hold newS
visit_cells(N, tr, tc, Wvec, tmp, 'W');
// check
if((int)ops_multi.size() + (int)tmp.size() > T) break;
// commit
ops_multi.insert(ops_multi.end(), tmp.begin(), tmp.end());
r=tr; c=tc; currentScore = newScore;
for(int id: Wvec) board[id] ^= curS;
}
// Compare multi vs baseline
if(currentScore > bestScore_single){
// multi-stage better
if((int)ops_multi.size() > T) ops_multi.resize(T);
for(char ch: ops_multi) cout<<ch<<'\n';
}else{
auto ops_base = build_baseline_ops(bestS_single);
for(char ch: ops_base) cout<<ch<<'\n';
}
return 0;
}
tnktsyk