結果

問題 No.2669 Generalized Hitting Set
ユーザー woodywoodywoodywoody
提出日時 2024-03-14 23:33:51
言語 C++14
(gcc 12.3.0 + boost 1.83.0)
結果
WA  
実行時間 -
コード長 5,948 bytes
コンパイル時間 4,937 ms
コンパイル使用メモリ 237,160 KB
実行使用メモリ 811,952 KB
最終ジャッジ日時 2024-03-14 23:34:02
合計ジャッジ時間 10,079 ms
ジャッジサーバーID
（参考情報） 		judge11 / judge13
このコードへのチャレンジ
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テストケース

テストケース表示
入力 結果 実行時間
実行使用メモリ
testcase_00 WA -
testcase_01 WA -
testcase_02 WA -
testcase_03 AC 253 ms
10,156 KB
testcase_04 AC 208 ms
8,912 KB
testcase_05 WA -
testcase_06 WA -
testcase_07 WA -
testcase_08 WA -
testcase_09 AC 194 ms
8,548 KB
testcase_10 WA -
testcase_11 WA -
testcase_12 AC 235 ms
9,760 KB
testcase_13 MLE -
testcase_14 -- -
testcase_15 -- -
testcase_16 -- -
testcase_17 -- -
testcase_18 -- -
testcase_19 -- -
testcase_20 -- -
testcase_21 -- -
testcase_22 -- -
testcase_23 -- -
testcase_24 -- -
testcase_25 -- -
testcase_26 -- -
testcase_27 -- -
testcase_28 -- -
testcase_29 -- -
testcase_30 -- -
testcase_31 -- -
testcase_32 -- -
testcase_33 -- -
testcase_34 -- -
testcase_35 -- -
testcase_36 -- -
testcase_37 -- -
testcase_38 -- -
testcase_39 -- -
testcase_40 -- -
権限があれば一括ダウンロードができます

ソースコード

diff # 

#include<bits/stdc++.h>
#include<atcoder/all>
#define rep(i,b) for(int i=0;i<b;i++)
#define rrep(i,b) for(int i=b-1;i>=0;i--)
#define rep1(i,b) for(int i=1;i<b;i++)
#define repx(i,x,b) for(int i=x;i<b;i++)
#define rrepx(i,x,b) for(int i=b-1;i>=x;i--)
#define fore(i,a) for(auto& i:a)
#define rng(x) (x).begin(), (x).end()
#define rrng(x) (x).rbegin(), (x).rend()
#define sz(x) ((int)(x).size())
#define pb push_back
#define fi first
#define se second
#define pcnt __builtin_popcountll

using namespace std;
using namespace atcoder;

using ll = long long;
using ld = long double;
template<typename T> using mpq = priority_queue<T, vector<T>, greater<T>>;
template<typename T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<typename T> bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
template<typename T> ll sumv(const vector<T>&a){ll res(0);for(auto&&x:a)res+=x;return res;}
bool yn(bool a) { if(a) {cout << "Yes" << endl; return true;} else {cout << "No" << endl; return false;}}
#define retval(x) {cout << #x << endl; return;}
#define cout2(x,y) cout << x << " " << y << endl;
#define coutp(p) cout << p.fi << " " << p.se << endl;
#define out cout << ans << endl;
#define outd cout << fixed << setprecision(20) << ans << endl;
#define outm cout << ans.val() << endl;
#define outv fore(yans , ans) cout << yans << "\n";
#define outdv fore(yans , ans) cout << yans.val() << "\n";
#define assertmle(x) if (!(x)) {vi v(3e8);}
#define asserttle(x) if (!(x)) {while(1){}}
#define coutv(v) {fore(vy , v) {cout << vy << " ";} cout << endl;}
#define coutv2(v) fore(vy , v) cout << vy << "\n";
#define coutvm(v) {fore(vy , v) {cout << vy.val() << " ";} cout << endl;}
#define coutvm2(v) fore(vy , v) cout << vy.val() << "\n";
using pll = pair<ll,ll>;using pil = pair<int,ll>;using pli = pair<ll,int>;using pii = pair<int,int>;using pdd = pair<ld,ld>;
using vi = vector<int>;using vd = vector<ld>;using vl = vector<ll>;using vs = vector<string>;using vb = vector<bool>;
using vpii = vector<pii>;using vpli = vector<pli>;using vpll = vector<pll>;using vpil = vector<pil>;
using vvi = vector<vector<int>>;using vvl = vector<vector<ll>>;using vvs = vector<vector<string>>;using vvb = vector<vector<bool>>;
using vvpii = vector<vector<pii>>;using vvpli = vector<vector<pli>>;using vvpll = vector<vpll>;using vvpil = vector<vpil>;
using mint = modint998244353;
//using mint = modint1000000007;
//using mint = dynamic_modint<0>;
using vm = vector<mint>;
using vvm = vector<vector<mint>>;
vector<int> dx={1,0,-1,0,1,1,-1,-1},dy={0,1,0,-1,1,-1,1,-1};
ll gcd(ll a, ll b) { return a?gcd(b%a,a):b;}
ll lcm(ll a, ll b) { return a/gcd(a,b)*b;}
#define yes {cout <<"Yes"<<endl;}
#define no {cout <<"No"<<endl;}
const double eps = 1e-10;
const ll LINF = 2001002003004005006ll;
const int INF = 1001001001;
#ifdef MY_LOCAL_DEBUG
#include "./debug/localDebug.cpp"
#define showp(p) cerr<<#p<<" = "<<p.fi<<" : "<<p.se<<endl
#define show1(a) cerr<<#a<<" = "<<a<<endl
#define show2(a,b) cerr<<#a<<" = "<<a<<" : "<<#b<<" = "<<b<<endl
#define show3(a,b,c) cerr<<#a<<" = "<<a<<" : "<<#b<<" = "<<b<<" : "<<#c<<" = "<<c<<endl
#define show4(a,b,c,d) cerr<<#a<<" = "<<a<<" : "<<#b<<" = "<<b<<" : "<<#c<<" = "<<c<<" : "<<#d<<" = "<<d<<endl
#define show5(a,b,c,d,e) cerr<<#a<<" = "<<a<<" : "<<#b<<" = "<<b<<" : "<<#c<<" = "<<c<<" : "<<#d<<" = "<<d<<" : "<<#e<<" = "<<e<<endl
#define DEBUG_LINE cout << "DEBUG_LINE : " << __LINE__ << endl
#define showv(v) {cout<<#v<<" : "; fore(vy , v) {cout << vy << " ";} cout << endl;}
#define showv2(v) {cout<<#v<<endl; fore(vy , v) cout << vy << "\n";}
#define showvm(v) {cout<<#v<<" : "; fore(vy , v) {cout << vy.val() << " ";} cout << endl;}
#define showvm2(v) {cout<<#v<<endl; fore(vy , v) cout << vy.val() << "\n";}
#define showmat(v) {cout<<#v<<endl; fore(row , v) { fore(seg , row) cout << seg << " "; cout << endl;}}
#define showmatm(v) {cout<<#v<<endl; fore(row , v) { fore(seg , row) cout << seg.val() << " "; cout << endl;}}
#else
#define showp(p)
#define show1(a)
#define show2(a,b)
#define show3(a,b,c)
#define show4(a,b,c,d)
#define show5(a,b,c,d,e)
#define DEBUG_LINE
#define showv(v)
#define showv2(v)
#define showvm(v)
#define showvm2(v)
#define showmat(v)
#define showmatm(v)
#endif
#define overload5(a,b,c,d,e,f, ...) f
#define show(...) overload5(__VA_ARGS__, show5, show4, show3, show2, show1)(__VA_ARGS__)

const int MAX_C = 25;
long long com[MAX_C][MAX_C];

// calculate nCk mod P in O(n^2) (1<=k<=n<=2000)
void cominit() {
    memset(com, 0, sizeof(com));
    com[0][0] = 1;
    for (int i = 1; i < MAX_C; ++i) {
        com[i][0] = 1;
        for (int j = 1; j < MAX_C; ++j) {
            com[i][j] = (com[i-1][j-1] + com[i-1][j]);
        }
    }
}

void solve(){
    int n,m,k; cin>>n>>m>>k;
    vs s(m);
    rep(i,m) cin>>s[i];

    cominit();

    vl coef(n+1);
    coef[k] = 1;
    repx(j,k+1,n+1){
        coef[j] = 1;
        rrepx(ii,k,j){
            coef[j] -= com[j][ii] * coef[ii];
        }
    }

    int l = 1 << n;

    vvl f_dp(n+1, vl(l));
    rep(i,m){
        int cnt = 0;
        rep(j,n) cnt += (1<<j) * (s[i][j]-'0');
        f_dp[0][cnt]++;
    }

    rep1(i,n+1){
        rep(j,l){
            f_dp[i][j] += f_dp[i-1][j];
            if ((j>>(i-1)&1) == 0) f_dp[i][j] += f_dp[i-1][j + (1 << (i-1))];
        }
    }
    vl f_sub(l);
    rep(i,l) f_sub[i] = f_dp[n][i] * coef[pcnt(i)];

    rep(i,n+1) rep(j,l) f_dp[i][j] = 0;
    f_dp[0] = f_sub;

    rep1(i,n+1){
        rep(j,l){
            f_dp[i][j] += f_dp[i-1][j];
            if (j>>(i-1)&1) f_dp[i][j] += f_dp[i-1][j - (1<<(i-1))];
        }
    }

    vl f = f_dp[n];
    vi mn(m+1,INF);
    rep(i,l){
        chmin(mn[f[i]], pcnt(i));
    }
    int ans = INF;
    rrep(i,m){
        chmin(ans, mn[i+1]);
        out;
    }

    return;
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);

    int t = 1;
    //cin>>t;

    rep(i,t){
        solve();
    }

    return 0;
}
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