【POJ 3076】 Sudoku
【题目链接】
http://poj.org/problem?id=3076
【算法】
将数独问题转化为精确覆盖问题,用Dancing Links求解
【代码】
#include <algorithm>
#include <bitset>
#include <cctype>
#include <cerrno>
#include <clocale>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <ctime>
#include <deque>
#include <exception>
#include <fstream>
#include <functional>
#include <limits>
#include <list>
#include <map>
#include <iomanip>
#include <ios>
#include <iosfwd>
#include <iostream>
#include <istream>
#include <ostream>
#include <queue>
#include <set>
#include <sstream>
#include <stdexcept>
#include <streambuf>
#include <string>
#include <utility>
#include <vector>
#include <cwchar>
#include <cwctype>
#include <stack>
#include <limits.h>
using namespace std;
#define MAXS 100000 struct info
{
int pos,val;
} a[MAXS]; int i,j,cnt;
int mat[][];
char s[]; inline int getRow(int pos)
{
return (pos - ) / + ;
}
inline int getCol(int pos)
{
return (pos - ) % + ;
}
inline int getGrid(int pos)
{
int x = getRow(pos),y = getCol(pos);
return (x - ) / * + (y - ) / + ;
}
struct DancingLinks
{
int n,m,step,size;
int U[MAXS],D[MAXS],L[MAXS],R[MAXS],Row[MAXS],Col[MAXS];
int H[MAXS],S[MAXS];
int ans[MAXS];
inline void init(int _n,int _m)
{
int i;
n = _n;
m = _m;
for (i = ; i <= m; i++)
{
S[i] = ;
U[i] = D[i] = i;
L[i] = i - ;
R[i] = i + ;
}
L[] = m; R[m] = ;
size = m;
for (i = ; i <= n; i++) H[i] = -;
}
inline void link(int r,int c)
{
size++;
Row[size] = r;
Col[size] = c;
S[c]++;
D[size] = D[c];
U[D[c]] = size;
U[size] = c;
D[c] = size;
if (H[r] < ) L[size] = R[size] = H[r] = size;
else
{
R[size] = R[H[r]];
L[R[H[r]]] = size;
L[size] = H[r];
R[H[r]] = size;
}
}
inline void Remove(int c)
{
int i,j;
R[L[c]] = R[c];
L[R[c]] = L[c];
for (i = D[c]; i != c; i = D[i])
{
for (j = R[i]; j != i; j = R[j])
{
D[U[j]] = D[j];
U[D[j]] = U[j];
S[Col[j]]--;
}
}
}
inline void Resume(int c)
{
int i,j;
for (i = U[c]; i != c; i = U[i])
{
for (j = L[i]; j != i; j = L[j])
{
D[U[j]] = j;
U[D[j]] = j;
S[Col[j]]++;
}
}
L[R[c]] = c;
R[L[c]] = c;
}
inline bool solve(int dep)
{
int i,j,c;
if (R[] == )
{
step = dep;
return true;
}
c = R[];
for (i = R[]; i != ; i = R[i])
{
if (S[i] < S[c])
c = i;
}
Remove(c);
for (i = D[c]; i != c; i = D[i])
{
ans[dep] = Row[i];
for (j = R[i]; j != i; j = R[j])
Remove(Col[j]);
if (solve(dep+)) return true;
for (j = L[i]; j != i; j = L[j])
Resume(Col[j]);
}
Resume(c);
return false;
}
} DLX; int main()
{ while (scanf("%s",s+) != EOF)
{
cnt = ;
memset(mat,,sizeof(mat));
for (i = ; i < ; i++) scanf("%s",s+i*+);
for (i = ; i <= ; i++)
{
if (s[i] != '-')
{
mat[][i] = ;
mat[][+(getRow(i)-)*+s[i]-'A'+] = ;
mat[][+(getCol(i)-)*+s[i]-'A'+] = ;
mat[][+(getGrid(i)-)*+s[i]-'A'+] = ;
} else
{
for (j = ; j <= ; j++)
{
cnt++;
mat[cnt][i] = ;
mat[cnt][+(getRow(i)-)*+j] = ;
mat[cnt][+(getCol(i)-)*+j] = ;
mat[cnt][+(getGrid(i)-)*+j] = ;
a[cnt] = (info){i,j};
}
}
}
DLX.init(cnt,);
for (i = ; i <= cnt; i++)
{
for (j = ; j <= ; j++)
{
if (mat[i][j])
DLX.link(i,j);
}
}
DLX.solve();
for (i = ; i < DLX.step; i++) s[a[DLX.ans[i]].pos] = 'A' + a[DLX.ans[i]].val - ;
for (i = ; i <= ; i++)
{
printf("%c",s[i]);
if (i % == ) printf("\n");
}
printf("\n");
} return ; }
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