16*16的数独， 跟9*9的类似。。。
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Sudoku
Description A Sudoku grid is a 16x16 grid of cells grouped in sixteen 4x4 squares, where some cells are filled with letters from A to P (the first 16 capital letters of the English alphabet), as shown in figure 1a. The game is to fill all the empty grid cells with letters from A to P such that each letter from the grid occurs once only in the line, the column, and the 4x4 square it occupies. The initial content of the grid satisfies the constraints mentioned above and guarantees a unique solution.
 Write a Sudoku playing program that reads data sets from a text file. Input Each data set encodes a grid and contains 16 strings on 16 consecutive lines as shown in figure 2. The i-th string stands for the i-th line of the grid, is 16 characters long, and starts from the first position of the line. String characters are from the set {A,B,…,P,-}, where – (minus) designates empty grid cells. The data sets are separated by single empty lines and terminate with an end of file.
Output The program prints the solution of the input encoded grids in the same format and order as used for input.
Sample Input --A----C-----O-I -J--A-B-P-CGF-H- --D--F-I-E----P- -G-EL-H----M-J-- ----E----C--G--- -I--K-GA-B---E-J D-GP--J-F----A-- -E---C-B--DP--O- E--F-M--D--L-K-A -C--------O-I-L- H-P-C--F-A--B--- ---G-OD---J----H K---J----H-A-P-L --B--P--E--K--A- -H--B--K--FI-C-- --F---C--D--H-N- Sample Output FPAHMJECNLBDKOGI OJMIANBDPKCGFLHE LNDKGFOIJEAHMBPC BGCELKHPOFIMAJDN MFHBELPOACKJGNID CILNKDGAHBMOPEFJ DOGPIHJMFNLECAKB JEKAFCNBGIDPLHOM EBOFPMIJDGHLNKCA NCJDHBAEKMOFIGLP HMPLCGKFIAENBDJO AKIGNODLBPJCEFMH KDEMJIFNCHGAOPBL GLBCDPMHEONKJIAF PHNOBALKMJFIDCEG IAFJOECGLDPBHMNK Source |
#include <stdio.h> #include <string.h> #include <iostream> using namespace std; #define N 300000 #define INF 0x3fffffff char g[20][20]; int ans[5000]; int u[N],d[N],r[N],l[N],num[N],H[N],save[N],save1[N]; int flag,head; const int n=4096; const int m=1024; int id; void prepare() { for(int i=0;i<=m;i++) { num[i]=0; d[i]=i; u[i]=i; r[i]=i+1; l[i+1]=i; } r[m]=0; memset(H,-1,sizeof(H)); // 记录每一行的第一个点 } void link(int tn,int tm) { id++; save1[id]=tn; // 记录行 ++num[save[id]=tm]; // 记录列 d[id]=d[tm]; u[ d[tm] ]=id; u[id]=tm; d[tm]=id; if(H[tn]<0) H[tn]=l[id]=r[id]=id; else { r[id]=r[H[tn]]; l[ r[H[tn]] ]=id; r[ H[tn] ]=id; l[id]=H[tn]; } } void build() { id=m; int sum; prepare(); int tn=0; for(int i=1;i<=256;i++) { for(int j=1;j<=16;j++) { ++tn; link(tn,i); } } sum=256; ///////////////// for(int i=1;i<=16;i++) // 每一行 { tn=(i-1)*256; for(int k=1;k<=16;k++) { int tk=tn+k; for(int j=1;j<=16;j++) { link(tk,sum+(i-1)*16+k); tk+=16; } } } sum+=256; /////////////////////// for(int i=1;i<=16;i++) { tn=(i-1)*16; for(int k=1;k<=16;k++) { int tk=tn+k; for(int j=1;j<=16;j++) { link(tk,sum+(i-1)*16+k); tk+=256; } } } sum+=256; ///////////////////////// int tt=0; for(int i1=1;i1<=4;i1++) { for(int j1=1;j1<=4;j1++) { tn=(i1-1)*256*4+16*4*(j1-1); for(int k=1;k<=16;k++) { ++tt; int tk; for(int i=1;i<=4;i++) { for(int j=1;j<=4;j++) { tk=tn+(i-1)*256+16*(j-1)+k; link(tk,sum+tt); } } } } } } void remove(int s) { l[ r[s] ]=l[s]; r[ l[s] ]=r[s]; for(int i=d[s];i!=s;i=d[i]) for(int j=r[i];j!=i;j=r[j]) { u[d[j]]=u[j]; d[u[j]]=d[j]; num[save[j]]--; } } void resume(int s) { r[l[s]]=s; l[r[s]]=s; for(int i=u[s];i!=s;i=u[i]) for(int j=l[i];j!=i;j=l[j]) { u[d[j]]=j; d[u[j]]=j; num[save[j]]++; } } void dfs(int s) { if(flag) return ; if(r[head]==head) { flag=1; for(int i=0;i<s;i++) { int ti,tj,tk; int tans=save1[ans[i]]-1; ti= (tans)/256+1; tj= (tans%256)/16+1; tk= (tans%256)%16+1; //printf("<%d %d> ",ti,tj); g[ti][tj]=tk+'A'-1; } return ; } int mi=INF,tu; for(int i=r[head];i!=head;i=r[i]) if(mi>num[i]) { mi=num[i]; tu=i; } remove(tu); for(int i=d[tu];i!=tu;i=d[i]) { for(int j=r[i];j!=i;j=r[j]) remove(save[j]); ans[s]=i; dfs(s+1); for(int j=l[i];j!=i;j=l[j]) resume(save[j]); } resume(tu); } int main() { while(scanf("%s",g[1]+1)!=EOF) { int tu=0; build(); for(int j=1;j<=16;j++) { if(g[1][j]!='-') { int kk=g[1][j]-'A'+1; remove( save[ H[tu+kk] ] ); for(int i1=r[ H[tu+kk] ];i1 != H[tu+kk];i1=r[i1]) { remove( save[i1] ); } } tu+=16; } for(int i=2;i<=16;i++) { scanf("%s",g[i]+1); for(int j=1;j<=16;j++) { if(g[i][j]!='-') { int kk=g[i][j]-'A'+1; remove( save[ H[tu+kk] ] ); for(int i1=r[ H[tu+kk] ];i1 != H[tu+kk];i1=r[i1]) { remove( save[i1] ); } } tu+=16; } } flag=0; dfs(0); for(int i=1;i<=16;i++) { for(int j=1;j<=16;j++) printf("%c",g[i][j]); printf("\n"); } printf("\n"); } return 0; }