p454.cpp
======================
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

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

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

int n, m;
vector<vector<string>> g;

void read() {
    cin >> n >> m;
    g.assign(n, vector<string>(m));
    for(int i = 0; i < n; i++) {
        for(int j = 0; j < m; j++) {
            cin >> g[i][j];
        }
    }
}

void solve() {
    // The grid is tiny (n, m <= 6), so this is brute-force backtracking over
    // white cells with sudoku-style pruning. For every black cell that carries
    // a clue we precompute the horizontal and vertical "runs" it controls: the
    // clue value, the length of the run, and a per-run id so each white cell
    // knows which two runs it participates in.
    //
    // The strong pruning step is to enumerate, for every run, the full list of
    // valid digit subsets ahead of time: every subset of {1..9} of size equal
    // to the run length whose digits sum to the clue. There are very few of
    // these per run, and they encode all the real constraints of the puzzle.
    //
    // During search we keep, for every run, a bitmask of digits already placed
    // in it. The set of digits still allowed in run r is then the union of all
    // valid subsets that are supersets of the current used mask, minus that
    // mask itself. For a white cell the candidate digits are the intersection
    // of its two runs' allowed sets, which is a much tighter filter than any
    // min/max-sum bound.
    //
    // On top of that we use MRV: at each step we pick the unfilled white cell
    // with the smallest candidate set, which is the standard sudoku
    // "most-constrained variable" heuristic and collapses the search tree on
    // the harder tests.

    vector<vector<int>> h_run(n, vector<int>(m, -1));
    vector<vector<int>> v_run(n, vector<int>(m, -1));
    vector<int> h_clue, v_clue;
    vector<int> h_count, v_count;

    for(int i = 0; i < n; i++) {
        for(int j = 0; j < m; j++) {
            const string& s = g[i][j];
            if(s[0] == '.' || s == "XXXXX") {
                continue;
            }

            string aa = s.substr(0, 2);
            string bb = s.substr(3, 2);

            if(bb != "XX") {
                int rid = (int)h_clue.size();
                h_clue.push_back(stoi(bb));
                int cnt = 0;
                for(int k = j + 1; k < m && g[i][k][0] == '.'; k++) {
                    h_run[i][k] = rid;
                    cnt++;
                }
                h_count.push_back(cnt);
            }

            if(aa != "XX") {
                int rid = (int)v_clue.size();
                v_clue.push_back(stoi(aa));
                int cnt = 0;
                for(int k = i + 1; k < n && g[k][j][0] == '.'; k++) {
                    v_run[k][j] = rid;
                    cnt++;
                }
                v_count.push_back(cnt);
            }
        }
    }

    vector<vector<vector<int>>> kak(10, vector<vector<int>>(46));
    for(int mask = 0; mask < (1 << 10); mask++) {
        if(mask & 1) {
            continue;
        }
        int cnt = __builtin_popcount(mask);
        int sum = 0;
        for(int x = 1; x <= 9; x++) {
            if(mask & (1 << x)) {
                sum += x;
            }
        }
        if(cnt <= 9 && sum <= 45) {
            kak[cnt][sum].push_back(mask);
        }
    }

    int H = (int)h_clue.size();
    int V = (int)v_clue.size();
    vector<vector<int>> h_subs(H), v_subs(V);
    for(int r = 0; r < H; r++) {
        h_subs[r] = kak[h_count[r]][h_clue[r]];
    }
    for(int r = 0; r < V; r++) {
        v_subs[r] = kak[v_count[r]][v_clue[r]];
    }

    vector<pair<int, int>> cells;
    for(int i = 0; i < n; i++) {
        for(int j = 0; j < m; j++) {
            if(g[i][j][0] == '.') {
                cells.push_back({i, j});
            }
        }
    }

    int total = (int)cells.size();
    vector<int> h_used(H, 0), v_used(V, 0);
    vector<bool> filled(total, false);
    vector<vector<int>> ans(n, vector<int>(m, 0));

    auto allowed_for = [](const vector<int>& subs, int used) {
        int allowed = 0;
        for(int t: subs) {
            if((t & used) == used) {
                allowed |= t;
            }
        }
        return allowed & ~used;
    };

    function<bool(int)> bt = [&](int placed) -> bool {
        if(placed == total) {
            return true;
        }

        vector<int> h_allowed(H), v_allowed(V);
        for(int r = 0; r < H; r++) {
            h_allowed[r] = allowed_for(h_subs[r], h_used[r]);
        }
        for(int r = 0; r < V; r++) {
            v_allowed[r] = allowed_for(v_subs[r], v_used[r]);
        }

        int best = -1, best_mask = 0, best_cnt = 100;
        for(int idx = 0; idx < total; idx++) {
            if(filled[idx]) {
                continue;
            }
            auto [i, j] = cells[idx];
            int cand = h_allowed[h_run[i][j]] & v_allowed[v_run[i][j]];
            int cnt = __builtin_popcount(cand);
            if(cnt < best_cnt) {
                best_cnt = cnt;
                best = idx;
                best_mask = cand;
                if(cnt <= 1) {
                    break;
                }
            }
        }

        if(best_cnt == 0) {
            return false;
        }

        auto [i, j] = cells[best];
        int hr = h_run[i][j], vr = v_run[i][j];
        filled[best] = true;

        for(int mask = best_mask; mask; mask &= mask - 1) {
            int bit = mask & -mask;
            int d = __builtin_ctz(bit);
            h_used[hr] |= bit;
            v_used[vr] |= bit;
            ans[i][j] = d;

            if(bt(placed + 1)) {
                return true;
            }

            h_used[hr] ^= bit;
            v_used[vr] ^= bit;
        }

        filled[best] = false;
        return false;
    };

    bt(0);

    for(int i = 0; i < n; i++) {
        for(int j = 0; j < m; j++) {
            if(j) {
                cout << ' ';
            }
            if(g[i][j][0] == '.') {
                cout << ans[i][j];
            } else {
                cout << '_';
            }
        }
        cout << '\n';
    }
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        solve();
    }

    return 0;
}

=================
p454.in1
======================
6 6
XXXXX XXXXX 28\XX 17\XX 28\XX XXXXX
XXXXX 22\22 ..... ..... ..... 10\XX
XX\34 ..... ..... ..... ..... .....
XX\14 ..... ..... 16\13 ..... .....
XX\22 ..... ..... ..... ..... XXXXX
XXXXX XX\16 ..... ..... XXXXX XXXXX

=================
statement.txt
======================
454. Kakuro
Time limit per test: 2.75 second(s)
Memory limit: 65536 kilobytes
input: standard
output: standard



Kakuro puzzle is played on a nx m grid of "black" and "white" cells. Apart from the top row and leftmost column which are entirely black, the grid has some amount of white cells which form "runs" and some amount of black cells. "Run" is a vertical or horizontal maximal one-lined block of adjacent white cells. Each row and column of the puzzle can contain more than one "run". Every white cell belongs to exactly two runs — one horizontal and one vertical run. Each horizontal "run" always has a number in the black half-cell to its immediate left, and each vertical "run" always has a number in the black half-cell immediately above it. These numbers are located in "black" cells and are called "clues".

The rules of the puzzle are simple:
place a single digit from 1 to 9 in each "white" cell
each digit may only be used once in each "run"
for all runs, the sum of all digits in a "run" must match the clue associated with the "run"


Given the grid, your task is to find a solution for the puzzle.


Picture of the first sample input
Picture of the first sample output


Input
The first line of input contains two integers n and m (2 ≤ n,m ≤ 6) — the number of rows and columns correspondingly. Each of the next n lines contains descriptions of m cells. Each cell description is one of the following 5-character strings:
..... — "white" cell;
XXXXX — "black" cell with no clues;
AA\BB — "black" cell with one or two clues. AA is either a 2-digit clue for the corresponding vertical run, or XX if there is no associated vertical run. BB is either a 2-digit clue for the corresponding horizontal run, or XX if there is no associated horizontal run.
The first row and the first column of the grid will never have any white cells. The given grid will have at least one "white" cell.

It is guaranteed that the given puzzle has at least one solution.

Output
Print n lines to the output with m cells in each line. For every "black" cell print '' (underscore), for every "white" cell print the corresponding digit from the solution. Delimit cells with a single space, so that each row consists of 2m-1 characters.

If there are many solutions, you may output any of them.

Example(s)
sample input
sample output
6 6
XXXXX XXXXX 28\XX 17\XX 28\XX XXXXX
XXXXX 22\22 ..... ..... ..... 10\XX
XX\34 ..... ..... ..... ..... .....
XX\14 ..... ..... 16\13 ..... .....
XX\22 ..... ..... ..... ..... XXXXX
XXXXX XX\16 ..... ..... XXXXX XXXXX
_ _ _ _ _ _
_ _ 5 8 9 _
_ 7 6 9 8 4
_ 6 8 _ 7 6
_ 9 2 7 4 _
_ _ 7 9 _ _

sample input
sample output
3 3
XXXXX 04\XX XXXXX
XX\04 ..... XXXXX
XXXXX XXXXX XXXXX
_ _ _
_ 4 _
_ _ _


=================
p454.in2
======================
3 3
XXXXX 04\XX XXXXX
XX\04 ..... XXXXX
XXXXX XXXXX XXXXX

=================
