"DLX algorithm" poj2676 Sudoku

The DLX algorithm solves the exact overlay problem template. You can see petition on the court.

#include <cstdio> #include <cstring> #include <vector>using namespace Std;const int sub[10][10]={{ 0,0,0,0,0,0,0,0,0,0},{0,1,1,1,2,2,2,3,3,3},{0,1,1,1,2,2,2,3,3,3},{0,1,1,1,2,2,2,3,3,3},{0,4,4,4,5,5,5,6,6,6},{ 0,4,4,4,5,5,5,6,6,6},{0,4,4,4,5,5,5,6,6,6},{0,7,7,7,8,8,8,9,9,9},{0,7,7,7,8,8,8,9,9,9},{0,7,7,7,8,8,8,9,9,9}}; int hlb[12],hub[12],llb[12],lub[12];const int maxn=4*9*9+5;const int Maxr=9*9*9+5;const int maxnode=9*9*9*4+maxn+5;// Ðð±àºå´ó1¿ªê¼£¬áð±àºåîª1~n£¬½áµã0êç±íí ½áµã£»½áµã1~nêç¸÷á𶥲¿µäðéäâ½áµãstruct DLX{int n,sz;//áðêý£¬½áµãxüêýint S [Maxn];//¸÷áð½áµãêýint row[maxnode],col[maxnode];//¸÷½áµãëùôúµäððáð±àºåint l[maxnode],r[maxnode],u[maxnode],d[ Maxnode];int ansd,ans[maxr];//½âvoid init (int n) {//nêçáðêýthis->n=n;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,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;} ˳xåá´±ía£¬±éàú³ýsíâµäæäëûôªëø #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;} BOOL Dfs (int d) {//printf ("%d", d); if (r[0]==0) {//õòµ½½âansd=d;//¼çâ¼½âµä³¤¶èreturn 1;} õò½áµãêýxîð¡µäáðc int C=r[0];//µúò»¸öî´é¾³ýµäáðfor (i,r,0) {if (S[i]<s[c]) {c=i;}} Remove (c);//é¾³ýµúcáðfor (i,d,c) {//óã½áµãiëùôúð𸲸çµúcáðans[d]=row[i]; for (j,r,i) {remove (col[j]);//é¾³ý½áµãiëùôúððäü¸²¸çµäëùóðæäëûáð}if (Dfs (d+1)) {return 1;} for (j,l,i) {restore (col[j]);//»ö¸´½áµãiëùôúððäü¸²¸çµäæäëûëùóðáð}}restore (c);//»ö¸´µúcáðreturn 0;} BOOL Solve (vector<int>& v) {v.clear (); if (!dfs (0)) {return 0;} for (int i=0;i<ansd;++i) {v.push_back (ans[i]);} return 1;}} Dlx;char s[12][12];int a[12][12];int Mah[1005],mal[1005],mav[1005];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;} int main () {int Zu;//freopen ("poj2676.in", "R", stdin),//freopen ("Poj2676.out", "w", stdout), memset (Hlb,0x7f,sizeof ( HLB)); Memset (Llb,0x7f,sizeof (LLB)); for (int. i=1;i<=9;++i) {for (int j=1;j<=9;++j) {hlb[sub[i][j]]=min (hlb[sub[ I][j]],i); Hub[sub[i][j]]=max (hub[sub[i][j]],i); Llb[sub[i][j]]=min (llb[sub[i][j]],j); Lub[sub[i][j]]=max (lub[sub I [J]],j);}} scanf ("%d", &zu), for (; zu;--zu) {for (int i=1;i<=9;++i) {scanf ("%s", s[i]+1);} int hang=0,lie=0;for (int i=1;i<=9;++i) {for (int j=1;j<=9;++j) {a[i][j]=s[i][j]-' 0 ';}} Dlx.init (9*9*4); for (int. i=1;i<=9;++i) {for (int. j=1;j<=9;++j) {for (int k=1;k<=9;++k) {if (!a[i][j] | | a[i][j]== k) {vector<int> Columns;columns.push_back (Encode (0,i-1,j-1)); Columns.push_back (Encode (1,i-1,k-1)); Columns.push_back (Encode (2,j-1,k-1)), Columns.push_back (Encode (3,sub[i][j]-1,k-1));d Lx.addrow (Encode (i-1,j-1,k-1), columns);}}} Vector<int> ans;dlx.solve (ANS); for (int i=0;i<ans.size (); ++i) {int r,c,v;decode (ANS[I],R,C,V); a[r+1][c+1]= v+1;} for (int i=1;i<=9;++i) {for (int j=1;j<9;++j) {printf ("%d", A[i][j]);} printf ("%d\n", A[i][9]);}} return 0;}

DLX algorithm poj2676 Sudoku