結果

問題 No.1045 直方体大学
ユーザー wk
提出日時 2020-05-03 17:38:48
言語 C++14
(gcc 13.3.0 + boost 1.87.0)
結果
WA  
実行時間 -
コード長 7,267 bytes
コンパイル時間 1,913 ms
コンパイル使用メモリ 170,840 KB
実行使用メモリ 18,928 KB
最終ジャッジ日時 2024-06-13 01:04:42
合計ジャッジ時間 3,015 ms
ジャッジサーバーID
（参考情報） 		judge1 / judge5
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 2
other AC * 5 WA * 12
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ソースコード

diff #
プレゼンテーションモードにする

#include <bits/stdc++.h>
#define REP(i, n) for(int i = 0; (i) < (n); (i)++)
using namespace std;
struct Edge{
    int to;
    long weight;
    Edge(int t, long w) : to(t), weight(w) {}
};
long modpow(long a, long n, long mod) {
    long res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}
// a^{-1} mod を計算する
long modinv(long a, long mod) {
    return modpow(a, mod - 2, mod);
}
struct compare1 { 
    bool operator()(const pair<long, long>& value,  
                                const long& key) 
    { 
        return (value.first < key); 
    } 
    bool operator()(const long& key,  
                    const pair<long, long>& value) 
    { 
        return (key < value.first); 
    } 
};
struct RMQ {
    vector<int> a;
    int inf = 2000000000; // 2*10^9
    int n = 1;
    RMQ(int n_ = 1){
        init(n_);
    }
    void init(int n_ = 1){
        while(n < n_) n *= 2;
        a.resize(2*n-1);
        REP(i, 2*n-1) a[i] = inf;
    }
    //k番目の値(0-indexed)をbに変更
    void update(int k, int b){
        k += n-1;
        a[k] = b;
        while(k > 0){
            k = (k-1)/2;
            a[k] = min(a[2*k+1], a[2*k+2]);
        }
    }
    //[c,b)の最小値を返す際に呼ぶ関数
    int query_first(int c, int b){
        return query(c, b, 0, 0, n);
    }
    //k : 節点番号, l, rはその接点が[l, r)に対応することを示す
    int query(int c, int b, int k, int l, int r){
        if(r <= c || b <= l) return inf;
        if(c <= l && r <= b) return a[k];
        else{
            int vl = query(c, b, k*2+1, l, (l+r)/2);
            int vr = query(c, b, k*2+2, (l+r)/2, r);
            return min(vl, vr);
        }
    }
};
struct UnionFind {
    vector<int> par;
    vector<int> rank;
    UnionFind(int n = 1){
        init(n);
    }
    void init(int n = 1){
        par.resize(n);
        rank.resize(n);
        REP(i, n) par[i] = i, rank[i] = 0; 
    }
    int root(int x){
        if(par[x] == x) return x;
        else return par[x] = root(par[x]);
    }
    bool issame(int x, int y){
        return root(x) == root(y);
    }
    bool merge(int x, int y){
        x = root(x); y = root(y);
        if(x == y) return false;
        if(rank[x] < rank[y]) swap(x, y);
        if(rank[x] == rank[y]) rank[x]++;
        par[y] = x;
        return true;
    }
};
template<class Abel> struct weightedUnionFind{
    vector<int> par;
    vector<int> rank;
    vector<Abel> diff_weight;
    weightedUnionFind(int n = 1, Abel SUM_UNITY = 0){
        init(n, SUM_UNITY);
    }
    void init(int n = 1, Abel SUM_UNITY = 0){
        par.resize(n); rank.resize(n); diff_weight.resize(n);
        REP(i, n) par[i] = i, rank[i] = 0, diff_weight[i] = SUM_UNITY; 
    }
    int root(int x){
        if(par[x] == x) return x;
        else{
            int r = root(par[x]);
            diff_weight[x] += diff_weight[par[x]];
            return par[x] = r;
        }
    }
    Abel weight(int x){
        root(x);
        return diff_weight[x];
    }
    bool issame(int x, int y){
        return root(x) == root(y);
    }
    bool merge(int x, int y, Abel w){
        w += weight(x); w -= weight(y);
        x = root(x); y = root(y);
        if(x == y) return false;
        if(rank[x] < rank[y]) swap(x, y), w = -w;
        if(rank[x] == rank[y]) rank[x]++;
        par[y] = x;
        diff_weight[y] = w;
        return true;
    }
    Abel diff(int x, int y){
        return weight(y) - weight(x);
    }
};
using Graph = vector<vector<int>>;
using P = pair<long, long>;
/*
void dijkstra(int s, int V, Graph &G, long* d){
    priority_queue<P, vector<P>, greater<P>> pque;
    fill(d, d + V, INF);
    d[s] = 0;
    pque.push(P(0, s));
    while(!pque.empty()){
        P p = pque.top(); pque.pop();
        int now = p.second;
        if(d[now] < p.first) continue;
        REP(i, G[now].size()){
            Edge e = G[now][i];
            if(d[e.to] > d[now] + e.weight){
                d[e.to] = d[now] + e.weight;
                pque.push(P(d[e.to], e.to));
            }
        }
    }
}
*/
int GCD(int a, int b){
    if(b == 0) return a; 
    if(a < b) return GCD(b, a);
    else return GCD(b, a%b);
}
struct BIT{
    vector<long> dat;
    int n = 1;
    BIT(int nn = 1){
        init(nn);
    }
    void init(int nn = 1){
        while(n < nn) n *= 2;
        dat.resize(n+1);
        REP(i, n+1) dat[i] = 0l;
    }
    //1-indexed!!!!
    //index iにx加える
    void add(int i, long x){
        while(i <= n){
            dat[i] += x;
            i += (i&(-i));
        }
    }
    //1-indexed!!!!!
    //index 1-iまでの和を求める
    long get_sum(int i){
        long ans = 0l;
        while(i > 0){
            ans += dat[i];
            i -= (i & (-i));
        }
        return ans;
    }
};
//{0, 1, 2, ..., n-1}までの中からk個の要素を持つ部分集合についての処理を行う
int next_combination(int sub){
    int x = sub & -sub, y = sub + x;
    return (((sub & ~y) / x) >> 1) | y;
}
//main関数内で
//bit = (1<<k)-1;
//for(; bit < (1<<n); bit = next_combination(bit))
// REP(i, n) if(bit & (1<<i)) でbitの中で選ばれている要素iを全部取得できる
//bitset<8>(bit)でbitを8桁の2進数で表示できる
int N;
int A[20], B[20], C[20];
int dp[(1<<16)+1][20][3];
int rec(int bit, int last, int men){
    //cout << "rec " << bit << " " << last << " " << men << endl;
    if(dp[bit][last][men] != -1) return dp[bit][last][men];
    if(bit == 0) return dp[bit][last][men] = 0;
    if(__builtin_popcount(bit) == 1 && bit != (1<<last)) return dp[bit][last][men] = 0;
    if(bit == (1<<last)){
        //cout << "equal " << bit << " " << last << " " << men << endl;
        if(men == 0) return dp[bit][last][men] = A[last];
        else if(men == 1) return dp[bit][last][men] = B[last];
        else return dp[bit][last][men] = C[last];
    }
    int ans = 0;
    int prev_bit = bit & ~(1<<last);
    int t, y, ta;
    if(men==0) ta = A[last], t = B[last], y = C[last];
    else if(men==1) t = A[last], ta = B[last], y = C[last];
    else t = A[last], y = B[last], ta = C[last];
    for(int u=0;u<N;u++){
        if(prev_bit & (1<<u)){
            for(int i=0;i<3;i++){
                int tate, yoko;
                if(i==0) tate = B[u], yoko = C[u];
                else if(i==1) tate = A[u], yoko = C[u];
                else tate = A[u], yoko = B[u];
                if((t>=tate && y>=yoko) || (t>=yoko && y>=tate)){
                    //cout << "tumu " << bit << " " << last << " " << men << " " << prev_bit << " " << u << " " << i << endl;
                    ans = max(ans, rec(prev_bit, u, i) + ta);
                }
                
            }
        }
    }
    return dp[bit][last][men] = ans;
}
int main()
{
    cin >> N;
    REP(i, N) cin >> A[i] >> B[i] >> C[i];
    REP(i, (1<<N)) REP(j, N) REP(k, 3) dp[i][j][k] = -1;
    int ans = 0;
    REP(i, N) REP(j, 3){
        //cout << rec((1<<N)-1, i, j) << endl;
        ans = max(ans, rec((1<<N)-1, i, j));
    }
    //REP(i, (1<<N)) REP(j, N) REP(k, 3) cout << i << " " << j << " " << k << " " << dp[i][j][k] << endl;
    cout << ans << endl;
    return 0;
}
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