#139876 (C++14) No.470 Inverse S+T Problem

• 問題: No.470 Inverse S+T Problem
• ユーザー: kimiyuki
• 提出日時: 2016-12-19 16:59:51
• 言語: C++14 (gcc 13.3.0 + boost 1.87.0)
• 結果: AC
• 実行時間: 18 ms / 2,000 ms
• コード長: 3,587 bytes
• コンパイル時間: 1,802 ms
• コンパイル使用メモリ: 109,088 KB
• 実行使用メモリ: 6,400 KB
• 最終ジャッジ日時: 2024-12-22 13:08:14
• 合計ジャッジ時間: 2,952 ms
• ジャッジサーバーID （参考情報）: judge5 / judge3

ファイルパターン	結果
sample	AC * 4
other	AC * 27

ソースコード

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
#include <iostream>
#include <vector>
#include <algorithm>
#include <map>
#include <cassert>
#define repeat(i,n) for (int i = 0; (i) < int(n); ++(i))
#define repeat_from(i,m,n) for (int i = (m); (i) < int(n); ++(i))
using namespace std;
struct strongly_connected_components {
    static pair<int,vector<int> > decompose(vector<vector<int> > const & g) { // adjacent list
        strongly_connected_components scc(g);
        return { scc.k, scc.c };
    }
private:
    int n;
    vector<vector<int> > to, from;
    explicit strongly_connected_components(vector<vector<int> > const & g) : n(g.size()), to(g), from(n) {
        repeat (i,n) for (int j : to[i]) from[j].push_back(i);
        decompose();
    }
    vector<bool> used;
    vector<int> vs;
    void dfs(int i) {
        used[i] = true;
        for (int j : to[i]) if (not used[j]) dfs(j);
        vs.push_back(i);
    }
    int k; // number of scc
    vector<int> c; // i-th vertex in g is in c_i-th vertex in scc-decomposed g
    void rdfs(int i) {
        used[i] = true;
        c[i] = k;
        for (int j : from[i]) if (not used[j]) rdfs(j);
    }
    void decompose() {
        used.clear(); used.resize(n, false);
        repeat (i,n) if (not used[i]) dfs(i);
        used.clear(); used.resize(n, false);
        k = 0;
        c.resize(n);
        reverse(vs.begin(), vs.end());
        for (int i : vs) if (not used[i]) {
            rdfs(i);
            k += 1;
        }
    }
};
vector<bool> twosat(int n, vector<pair<int, int> > const & cnf) {
    vector<vector<int> > g(2*n);
    auto i = [&](int x) { assert (x != 0 and abs(x) <= n); return x > 0 ? x-1 : n-x-1; };
    for (auto it : cnf) {
        int x, y; tie(x, y) = it; // x or y
        g[i(- x)].push_back(i(y)); // not x implies y
        g[i(- y)].push_back(i(x)); // not y implies x
    }
    vector<int> component = strongly_connected_components::decompose(g).second;
    vector<bool> valuation(n);
    repeat_from (x,1,n+1) {
        if (component[i(x)] == component[i(- x)]) { // x iff not x
            return vector<bool>(); // unsat
        }
        valuation[x-1] = component[i(x)] > component[i(- x)]; // use components which indices are large
    }
    return valuation;
}
int main() {
    int n; cin >> n;
    vector<string> s(n); repeat (i,n) cin >> s[i];
    assert (1 <= n and n <= 100000);
    repeat (i,n) {
        assert (s[i].length() == 3);
        for (char c : s[i]) assert ('A' <= c and c <= 'Z' or 'a' <= c and c <= 'z');
    }
    vector<bool> result;
    if (n <= 52) {
        //     x_i : U_i = S + TT
        // not x_i : U_i = SS + T
        vector<pair<int, int> > cnf;
        map<string, vector<int> > used;
        repeat (i,n) {
            int x = i + 1;
            used[s[i].substr(0, 1)].push_back(+ x);
            used[s[i].substr(1, 2)].push_back(+ x);
            used[s[i].substr(0, 2)].push_back(- x);
            used[s[i].substr(2, 1)].push_back(- x);
        }
        for (auto it : used) {
            for (int x : it.second) for (int y : it.second) if (x < y) {
                // cerr << "not " << x << " or " << "not " << y << endl;
                cnf.emplace_back(- x, - y); // not x or not y
            }
        }
        result = twosat(n, cnf);
    }
    if (result.empty()) {
        cout << "Impossible" << endl;
    } else {
        repeat (i,n) {
            if (result[i]) {
                cout << s[i][0] << ' ' << s[i][1] << s[i][2] << endl;
            } else {
                cout << s[i][0] << s[i][1] << ' ' << s[i][2] << endl;
            }
        }
    }
    return 0;
}
 
 
הההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההה
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX