UVa 1461 - Sudoku Extension

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链接

传送门

题意

给出一个数独，其中0可以填1到9任意数字，‘e’只能填偶数，‘o’只能填奇数，其余字母相同字母的格子的数字必须相同。输出给出的数独有多少个解。

思路

DLX基础题，注意限制。

代码

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#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cctype>

using namespace std;

const int maxr = 750;
const int maxn = 330;
const int maxnode = 20000;

// 行编号从1开始，列编号为1~n，结点0是表头结点; 结点1~n是各列顶部的虚拟结点
struct DLX {
  int n, sz; // 列数，结点总数
  int S[maxn]; // 各列结点数
  
  int row[maxnode], col[maxnode]; // 各结点行列编号
  int L[maxnode], R[maxnode], U[maxnode], D[maxnode]; // 十字链表
  
  int ansd, ans[maxr]; // 解
  
  void init(int n) { // n是列数
    this->n = n;
    // 虚拟结点
    for(int i = 0 ; i <= n; ++i) {
      U[i] = i; D[i] = i; L[i] = i-1, R[i] = i+1;
    }
    R[n] = 0; L[0] = n;
    sz = n + 1;
    memset(S, 0, sizeof(S));
  }
  
  void addRow(int r, const vector<int>& columns) {
    int first = sz;
    for(int i = 0; i < columns.size(); ++i) {
      int c = columns[i];
      L[sz] = sz - 1; R[sz] = sz + 1; D[sz] = c; U[sz] = U[c];
      D[U[c]] = sz; U[c] = sz;
      row[sz] = r; col[sz] = c;
      S[c]++; sz++;
    }
    R[sz - 1] = first; L[first] = sz - 1;
  }
  
  // 顺着链表A，遍历除s外的其他元素
#define FOR(i,A,s) for(int i = A[s]; i != s; i = A[i])
  
  void remove(int c) {
    L[R[c]] = L[c];
    R[L[c]] = R[c];
    FOR(i,D,c)
    FOR(j,R,i) { U[D[j]] = U[j]; D[U[j]] = D[j]; --S[col[j]]; }
  }
  
  void restore(int c) {
    FOR(i,U,c)
    FOR(j,L,i) { ++S[col[j]]; U[D[j]] = j; D[U[j]] = j; }
    L[R[c]] = c;
    R[L[c]] = c;
  }
  
  int dfs(int d) {
    if (R[0] == 0) {
      ansd = d;
      return 1;
    }
    int c = R[0], res = 0;
    FOR(i,R,0) if(S[i] < S[c]) c = i;
    remove(c);
    FOR(i,D,c) {
      ans[d] = row[i];
      FOR(j,R,i) remove(col[j]);
      res += dfs(d + 1);
      FOR(j,L,i) restore(col[j]);
    }
    restore(c);
    return res;
  }
  
  bool solve(vector<int>& v) {
    v.clear();
    if(!dfs(0)) return false;
    for(int i = 0; i < ansd; i++) v.push_back(ans[i]);
    return true;
  }
  
} solver;

const int SLOT = 0;
const int ROW = 1;
const int COL = 2;
const int SUB = 3;

int encode(int a, int b, int c) {
  return a * 81 + b * 9 + c + 1;
}

void decode(int code, int& a, int& b, int& c) {
  --code;
  c = code % 9; code /= 9;
  b = code % 9; code /= 9;
  a = code;
}

typedef pair<int, int> pii;

char s[15][15];
vector<pii> a[27];

int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    for (int i = 0; i < 9; ++i) {
      scanf("%s", s[i]);
    }
    fill(a, a + 26, vector<pii>());
    solver.init(324);
    for (int r = 0; r < 9; ++r) {
      for (int c = 0; c < 9; ++c) {
        if (isdigit(s[r][c])) {
          for (int v = 0; v < 9; ++v) {
            if (s[r][c] == '0' || s[r][c] == v + '1') {
              vector<int> columns;
              columns.push_back(encode(SLOT, r, c));
              columns.push_back(encode(ROW, r, v));
              columns.push_back(encode(COL, c, v));
              columns.push_back(encode(SUB, (r / 3) * 3 + c / 3, v));
              solver.addRow(encode(r, c, v), columns);
            }
          }
        } else if (s[r][c] == 'e' || s[r][c] == 'o') {
          for (int v = s[r][c] == 'e' ? 1 : 0; v < 9; v += 2) {
            vector<int> columns;
            columns.push_back(encode(SLOT, r, c));
            columns.push_back(encode(ROW, r, v));
            columns.push_back(encode(COL, c, v));
            columns.push_back(encode(SUB, (r / 3) * 3 + c / 3, v));
            solver.addRow(encode(r, c, v), columns);
          }
        } else {
          a[s[r][c] - 'a'].push_back(make_pair(r, c));
        }
      }
    }
    for (int i = 0; i < 26; ++i) {
      if (a[i].empty()) {
        continue;
      }
      int r = 0, c = 0;
      for (int v = 0; v < 9; ++v) {
        vector<int> columns;
        for (int j = 0; j < a[i].size(); ++j) {
          r = a[i][j].first, c = a[i][j].second;
          columns.push_back(encode(SLOT, r, c));
          columns.push_back(encode(ROW, r, v));
          columns.push_back(encode(COL, c, v));
          columns.push_back(encode(SUB, (r / 3) * 3 + c / 3, v));
        }
        solver.addRow(encode(r, c, v), columns);
      }
    }
    printf("%d\n", solver.dfs(0));
  }
  return 0;
}