結果
| 問題 | No.2278 Time Bomb Game 2 |
| ユーザー |
 Ujjwal Rana
|
| 提出日時 | 2023-04-21 22:45:54 |
| 言語 | C++17(gcc12) (gcc 12.3.0 + boost 1.87.0) |
| 結果 |
AC
|
| 実行時間 | 28 ms / 2,000 ms |
| コード長 | 7,073 bytes |
| コンパイル時間 | 16,255 ms |
| コンパイル使用メモリ | 321,772 KB |
| 最終ジャッジ日時 | 2025-02-12 12:18:26 |
|
ジャッジサーバーID （参考情報） |
judge1 / judge5 |
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| ファイルパターン | 結果 |
|---|---|
| sample | AC * 2 |
| other | AC * 70 |
コンパイルメッセージ
main.cpp: In function 'bool solve()':
main.cpp:276:16: warning: 'okg' may be used uninitialized [-Wmaybe-uninitialized]
276 | a=t-abs(k-okg);
| ~~~^~~~~~~
main.cpp:239:8: note: 'okg' was declared here
239 | ll okg,ind;
| ^~~
ソースコード
#pragma GCC target ("avx2")#pragma GCC optimize ("O3")#pragma GCC optimize("Ofast")#pragma GCC optimize ("unroll-loops")// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")#include <bits/stdc++.h>#include <ext/pb_ds/assoc_container.hpp>using namespace __gnu_pbds;using namespace std;#define io ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)#define all(x) (x).begin(), (x).end()#define rall(x) (x).rbegin(), (x).rend()#define endl '\n'typedef long long ll;#define mod1 (ll)1000000007#define mod2 (ll)998244353#define pll pair<ll,ll>typedef long double lb;typedef tree<pair<ll, ll>,null_type,less<pair<ll, ll>>,rb_tree_tag,tree_order_statistics_node_update> ordered_set;#define eps (lb)(1e-9)struct custom_hash {static uint64_t splitmix64(uint64_t x) {x += 0x9e3779b97f4a7c15;x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;x = (x ^ (x >> 27)) * 0x94d049bb133111eb;return x ^ (x >> 31);}size_t operator()(uint64_t x) const {static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();return splitmix64(x + FIXED_RANDOM);}};// Operator overloadstemplate<typename T1, typename T2> // cin >> pair<T1, T2>istream& operator>>(istream &istream, pair<T1, T2> &p) { return (istream >> p.first >> p.second); }template<typename T> // cin >> vector<T>istream& operator>>(istream &istream, vector<T> &v){for (auto &it : v)cin >> it;return istream;}template<typename T1, typename T2> // cout << pair<T1, T2>ostream& operator<<(ostream &ostream, const pair<T1, T2> &p) { return (ostream << p.first << " " << p.second); }template<typename T> // cout << vector<T>ostream& operator<<(ostream &ostream, const vector<T> &c) { for (auto &it : c) cout << it << " "; return ostream; }// Utility functionstemplate <typename T>void print(T &&t) { cout << t << "\n"; }template <typename T, typename... Args>void print(T &&t, Args &&... args){cout << t << " ";print(forward<Args>(args)...);}mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());ll random(ll p){ // gives random number in [0,p]return uniform_int_distribution<ll>(0, p)(rng);}class DisjSet {vector<int>sz,parent; int n;public:int connected;// Constructor to create and// initialize sets of n itemsDisjSet(int n){sz = vector<int>(n,1);parent = sz;this->n = n;this->connected=n;makeSet();}// Creates n single item setsvoid makeSet(){for (int i = 0; i < n; i++) {parent[i] = i;}}// Finds set of given item xint find(int x){// Finds the representative of the set// that x is an element ofif (parent[x] != x) {// if x is not the parent of itself// Then x is not the representative of// his set,parent[x] = find(parent[x]);// so we recursively call Find on its parent// and move i's node directly under the// representative of this set}return parent[x];}int size(int x){return sz[find(x)];}// Do union of two sets represented// by x and y.void Union(int x, int y){// Find current sets of x and yint xset = find(x);int yset = find(y);// If they are already in same setif (xset == yset){return;}connected--;// Put smaller ranked item under// bigger ranked item if ranks are// differentif (sz[xset] <= sz[yset]) {parent[xset] = yset;sz[yset] += sz[xset];}else {parent[yset] = xset;sz[xset] += sz[yset];}}};DisjSet dsu(2);ll n,k,t;vector<ll>v,las;ll calc(ll ind,ll moves){ // it's on border and connected with other component// cout<<ind<<" "<<moves<<endl;if(moves<=0){return v[ind];}else if(moves==1){return v[ind]^1;}else if(moves%2==0 and dsu.size(ind)%2){return v[ind];}else if(dsu.size(ind)>1 and (moves%2)){return v[ind]^1;}else if(dsu.size(ind)==1){if((ind+1)<n){if(calc(ind+1,moves-1)!=v[ind]){return v[ind]^1;}}if((ind-1)>=0){if(calc(ind-1,moves-1)!=v[ind]){return v[ind]^1;}}return v[ind];}else if(dsu.find(ind)==dsu.find(0)){return v[ind];}else if(dsu.find(ind)==dsu.find(n-1)){return v[ind];}else if(v[ind-1]!=v[ind]){ll k=ind-1;for(ll j(ind);j<n;++j){if(dsu.find(j)!=dsu.find(ind)){k=j;break;}}if(moves>=(k-ind)){return v[ind]^1;}else{return v[ind];}}else{ll k=ind+1;for(ll j(ind);j>-1;--j){if(dsu.find(j)!=dsu.find(ind)){k=j;break;}}if(moves>=(ind-k)){return v[ind]^1;}else{return v[ind];}}}bool solve();int main() {io;ll t=1,n=1;// cin>>t;while (t--){solve()?cout<<"Alice"<<endl:cout<<"Bob"<<endl;;}return 0;}bool solve(){cin>>n>>k>>t;v.resize(n);las=v;dsu=DisjSet(n);for(ll i(0);i<n;++i){char p;cin>>p;ll t=(p-'A');v[i]=t;if(i>0 and v[i]==v[i-1]){dsu.Union(i,i-1);}}ll a=t,b=t;ll okg,ind;if(dsu.connected==1){return v[0];}--k;if(dsu.find(n-1)==dsu.find(k)){for(ll t(k);t>-1;--t){if(dsu.find(k)!=dsu.find(t)){okg=t; break;}}okg++;a=t-abs(k-okg);ind=okg;if(calc(okg,a)!=v[ind]){return v[ind]^1;}return v[ind];}else if(dsu.find(0)==dsu.find(k)){for(ll t(k);t<n;++t){if(dsu.find(k)!=dsu.find(t)){okg=t; break;}}okg--;a=t-abs(k-okg);ind=okg;if(calc(okg,a)!=v[ind]){return v[ind]^1;}return v[ind];}else{for(ll t(k);t<n;++t){if(dsu.find(k)!=dsu.find(t)){okg=t; break;}}okg--;a=t-abs(k-okg);ind=okg;if(calc(okg,a)!=v[ind]){return v[ind]^1;}for(ll t(k);t>-1;--t){if(dsu.find(k)!=dsu.find(t)){okg=t; break;}}okg++;a=t-abs(k-okg);ind=okg;if(calc(okg,a)!=v[ind]){return v[ind]^1;}return v[ind];}}// Do not get stuck on a single approach for long, think of multiple ways