UVa 1461 - Sudoku Extension

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;
}