Submission #384814

#	Time	Username	Problem	Language	Result	Execution time	Memory
384814		ACmachine	Lamps (JOI19_lamps)	C++17	100 / 100	413 ms	199932 KiB

lamp

#include <bits/stdc++.h>
using namespace std;
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;

template<typename T> using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template<typename T, typename U> using ordered_map = tree<T, U, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

typedef long long ll;
typedef long double ld;

typedef pair<int,int> pii;
typedef pair<ll,ll> pll;

typedef vector<int> vi;
typedef vector<ll> vll;

#define FOR(i,j,k,in) for(int i=(j); i < (k);i+=in)
#define FORD(i,j,k,in) for(int i=(j); i >=(k);i-=in)
#define REP(i,b) FOR(i,0,b,1)
#define REPD(i,b) FORD(i,b,0,1)
#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define all(x) begin(x), end(x)
#define MANY_TESTS int tcase; cin >> tcase; while(tcase--)

const double EPS = 1e-9;
const int MOD = 1e9+7;
const ll INFF = 1e18;
const int INF = 1e9;
const ld PI = acos((ld)-1);
const vi dy = {1, 0, -1, 0, -1, 1, 1, -1};
const vi dx = {0, 1, 0, -1, -1, 1, -1, 1};

void DBG(){cout << "]" << endl;}
template<typename T, typename ...U> void DBG(const T& head, const U... args){ cout << head << "; "; DBG(args...); }
#define dbg(...) cout << "Line(" << __LINE__ << ") -> [" << #__VA_ARGS__ << "]: [", DBG(__VA_ARGS__);
#define chk() cout << "Check at line(" << __LINE__ << ") hit." << endl;

template<class T, unsigned int U>
ostream& operator<<(ostream& out, const array<T, U> &v){out << "[";  REP(i, U) out << v[i] << ", ";  out << "]"; return out;}
template <class T, class U>
ostream& operator<<(ostream& out, const pair<T, U> &par) {out << "[" << par.first << ";" << par.second << "]"; return out;}
template <class T>
ostream& operator<<(ostream& out, const set<T> &cont) { out << "{"; for( const auto &x:cont) out << x << ", "; out << "}"; return out; }
template <class T, class U>
ostream& operator<<(ostream& out, const map<T, U> &cont) {out << "{"; for( const auto &x:cont) out << x << ", "; out << "}"; return out; }
template<class T>
ostream& operator<<(ostream& out, const vector<T> &v){ out << "[";  REP(i, v.size()) out << v[i] << ", ";  out << "]"; return out;}

template<class T>
istream& operator>>(istream& in, vector<T> &v){ for(auto &x : v) in >> x; return in; }

int main(){
 	ios_base::sync_with_stdio(false);
 	cin.tie(NULL); cout.tie(NULL);
    int n; cin >> n;
    string a, b;
    cin >> a >> b;
    vector<vector<vector<int>>> dp(n + 1, vector<vector<int>>(3, vector<int>(2, INF))); // ofc always toggle before set
    // so like last sol, maybe I won't make mistake in impl
    dp[0][2][0] = 0;
    dp[0][2][1] = 1;
    REP(i, n){
        REP(setting, 3){
            dp[i][setting][0] = min(dp[i][setting][0], dp[i][setting][1]);
            dp[i][setting][1] = min(dp[i][setting][1], dp[i][setting][0] + 1); // like in the prev sol, swap / no swap
            REP(swapping, 2){
                // setting has priority
                int curr = a[i] - '0';
                int oth = b[i] - '0';
                // not keep setting
                if((curr ^ swapping) == oth){
                    dp[i + 1][2][swapping] = min(dp[i + 1][2][swapping], dp[i][setting][swapping]);
                }
                // keep setting
                int ncurr = curr;
                if(setting != 2)
                    ncurr = setting;
                if((ncurr ^ swapping) == oth){
                    dp[i + 1][setting][swapping] = min(dp[i + 1][setting][swapping], dp[i][setting][swapping]);
                }
                // set to something because not equal
                dp[i + 1][oth ^ swapping][swapping] = min(dp[i + 1][oth ^ swapping][swapping], dp[i][setting][swapping] + 1);
            }
        }
    }
    int ans = INF;
    REP(setting, 3){
        REP(swapping, 2){
            ans = min(ans, dp[n][setting][swapping]);
        }
    }
    cout << ans << "\n";
    return 0;
}

• Subtask 1: 6 / 6
• Subtask 2: 41 / 41
• Subtask 3: 4 / 4
• Subtask 4: 49 / 49