POJ 3074 :
DescriptionIn the game of Sudoku, you are given a large 9 × 9 grid divided into smaller 3 × 3 subgrids. For example,
. 2 7 3 8 . . 1 . . 1 . . . 6 7 3 5 . . . . . . . 2 9 3 . 5 6 9 2 . 8 . . . . . . . . . . . 6 . 1 7 4 5 . 3 6 4 . . . . . . . 9 5 1 8 . . . 7 . . 8 . . 6 5 3 4 . Given some of the numbers in the grid, your goal is to determine the remaining numbers such that the numbers 1 through 9 appear exactly once in (1) each of nine 3 × 3 subgrids, (2) each of the nine rows, and (3) each of the nine columns.
Input
The input test file will contain multiple cases. Each test case consists of a single line containing 81 characters, which represent the 81 squares of the Sudoku grid, given one row at a time. Each character is either a digit (from 1 to 9) or a period (used to indicate an unfilled square). You may assume that each puzzle in the input will have exactly one solution. The end-of-file is denoted by a single line containing the word “end”.
Output
For each test case, print a line representing the completed Sudoku puzzle.
Sample Input
.2738..1..1...6735.......293.5692.8...........6.1745.364.......9518...7..8..6534. ......52..8.4......3...9...5.1...6..2..7........3.....6...1..........7.4.......3. endSample Output
527389416819426735436751829375692184194538267268174593643217958951843672782965341 416837529982465371735129468571298643293746185864351297647913852359682714128574936
POJ 3076:
DescriptionA 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
这两道题几乎一样的,就是要你求一个数独矩阵。
难得有这样一道接近生活的信息题啊~~~
POJ 3074:
1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 using namespace std; 5 const int maxnode=100010; 6 const int maxn=755; 7 const int maxm=345; 8 struct DLX 9 { 10 int L[maxnode],R[maxnode],U[maxnode],D[maxnode],Row[maxnode],Col[maxnode],C[maxm],H[maxn],cnt; 11 bool used[maxn]; 12 void Init(int n,int m) 13 { 14 for(int i=0;i<=m;i++) 15 { 16 L[i]=i-1;R[i]=i+1; 17 U[i]=D[i]=i;C[i]=0; 18 } 19 cnt=m;L[0]=m;R[m]=0; 20 21 for(int i=1;i<=n;i++) 22 H[i]=0,used[i]=false; 23 } 24 void Link(int x,int y) 25 { 26 C[Col[++cnt]=y]++; 27 Row[cnt]=x; 28 29 U[cnt]=y; 30 U[D[y]]=cnt; 31 D[cnt]=D[y]; 32 D[y]=cnt; 33 34 if(H[x]) 35 L[R[H[x]]]=cnt,R[cnt]=R[H[x]],R[H[x]]=cnt,L[cnt]=H[x]; 36 else 37 H[x]=L[cnt]=R[cnt]=cnt; 38 } 39 40 void Delete(int c) 41 { 42 L[R[c]]=L[c];R[L[c]]=R[c]; 43 for(int i=D[c];i!=c;i=D[i]) 44 for(int j=R[i];j!=i;j=R[j]) 45 --C[Col[j]],U[D[j]]=U[j],D[U[j]]=D[j]; 46 } 47 48 void Resume(int c) 49 { 50 L[R[c]]=c;R[L[c]]=c; 51 for(int i=U[c];i!=c;i=U[i]) 52 for(int j=L[i];j!=i;j=L[j]) 53 ++C[Col[j]],U[D[j]]=j,D[U[j]]=j; 54 } 55 56 bool Solve() 57 { 58 if(!R[0])return true; 59 int p=R[0]; 60 for(int i=R[p];i;i=R[i]) 61 if(C[p]>C[i]) 62 p=i; 63 Delete(p); 64 for(int i=D[p];i!=p;i=D[i]){ 65 used[Row[i]]=true; 66 for(int j=R[i];j!=i;j=R[j]) 67 Delete(Col[j]); 68 if(Solve()) 69 return true; 70 used[Row[i]]=false; 71 for(int j=L[i];j!=i;j=L[j]) 72 Resume(Col[j]); 73 } 74 Resume(p); 75 return false; 76 } 77 void Print() 78 { 79 for(int i=1;i<=81;i++) 80 for(int j=(i-1)*9+1;j<=i*9;j++) 81 if(used[j]){ 82 int Color=j-(i-1)*9; 83 printf("%d",Color); 84 } 85 printf(" "); 86 } 87 }DLX; 88 89 int Area(int x,int y) 90 { 91 if(x<=3&&y<=3)return 0; 92 if(x<=3&&y<=6)return 1; 93 if(x<=3)return 2; 94 if(x<=6&&y<=3)return 3; 95 if(x<=6&&y<=6)return 4; 96 if(x<=6)return 5; 97 if(y<=3)return 6; 98 if(y<=6)return 7; 99 return 8; 100 } 101 102 char str[110]; 103 int main() 104 { 105 int x,y; 106 while(~scanf("%s",str+1)) 107 { 108 if(!strcmp(str+1,"end"))break; 109 DLX.Init(729,324);x=1;y=1; 110 for(int i=1;i<=81;i++) 111 { 112 for(int j=(i-1)*9+1;j<=i*9;j++) 113 { 114 int Color=j-(i-1)*9; 115 if(str[i]!='.'&&str[i]-'0'!=Color) 116 continue; 117 118 DLX.Link(j,(x-1)*9+Color); //行中对应颜色 119 DLX.Link(j,81+(y-1)*9+Color); //列中对应颜色 120 DLX.Link(j,162+Area(x,y)*9+Color);//块中对应颜色 121 DLX.Link(j,243+i); //矩阵中对应位置 122 } 123 y++;x+=y/10;y=(y-1)%9+1; 124 } 125 DLX.Solve(); 126 DLX.Print(); 127 } 128 return 0; 129 }
POJ 3076:
1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 using namespace std; 5 const int maxnode=500010; 6 const int maxn=4100; 7 const int maxm=1030; 8 struct DLX 9 { 10 int L[maxnode],R[maxnode],U[maxnode],D[maxnode],Row[maxnode],Col[maxnode],C[maxm],H[maxn],cnt; 11 bool used[maxn]; 12 void Init(int n,int m) 13 { 14 for(int i=0;i<=m;i++) 15 { 16 L[i]=i-1;R[i]=i+1; 17 U[i]=D[i]=i;C[i]=0; 18 } 19 cnt=m;L[0]=m;R[m]=0; 20 21 for(int i=1;i<=n;i++) 22 H[i]=0,used[i]=false; 23 } 24 void Link(int x,int y) 25 { 26 C[Col[++cnt]=y]++; 27 Row[cnt]=x; 28 29 U[cnt]=y; 30 U[D[y]]=cnt; 31 D[cnt]=D[y]; 32 D[y]=cnt; 33 34 if(H[x]) 35 L[R[H[x]]]=cnt,R[cnt]=R[H[x]],R[H[x]]=cnt,L[cnt]=H[x]; 36 else 37 H[x]=L[cnt]=R[cnt]=cnt; 38 } 39 40 void Delete(int c) 41 { 42 L[R[c]]=L[c];R[L[c]]=R[c]; 43 for(int i=D[c];i!=c;i=D[i]) 44 for(int j=R[i];j!=i;j=R[j]) 45 --C[Col[j]],U[D[j]]=U[j],D[U[j]]=D[j]; 46 } 47 48 void Resume(int c) 49 { 50 L[R[c]]=c;R[L[c]]=c; 51 for(int i=U[c];i!=c;i=U[i]) 52 for(int j=L[i];j!=i;j=L[j]) 53 ++C[Col[j]],U[D[j]]=j,D[U[j]]=j; 54 } 55 56 bool Solve() 57 { 58 if(!R[0])return true; 59 int p=R[0]; 60 for(int i=R[p];i;i=R[i]) 61 if(C[p]>C[i]) 62 p=i; 63 Delete(p); 64 for(int i=D[p];i!=p;i=D[i]){ 65 used[Row[i]]=true; 66 for(int j=R[i];j!=i;j=R[j]) 67 Delete(Col[j]); 68 if(Solve()) 69 return true; 70 used[Row[i]]=false; 71 for(int j=L[i];j!=i;j=L[j]) 72 Resume(Col[j]); 73 } 74 Resume(p); 75 return false; 76 } 77 void Print() 78 { 79 for(int i=1;i<=256;i++){ 80 for(int j=(i-1)*16+1;j<=i*16;j++) 81 if(used[j]){ 82 int Color=j-(i-1)*16; 83 printf("%c",'A'+Color-1); 84 break; 85 } 86 if(i%16==0) 87 printf(" "); 88 } 89 printf(" "); 90 } 91 }DLX; 92 93 int Area(int x,int y) 94 { 95 if(x<=4&&y<=4)return 0; 96 if(x<=4&&y<=8)return 1; 97 if(x<=4&&y<=12)return 2; 98 if(x<=4)return 3; 99 100 if(x<=8&&y<=4)return 4; 101 if(x<=8&&y<=8)return 5; 102 if(x<=8&&y<=12)return 6; 103 if(x<=8)return 7; 104 105 if(x<=12&&y<=4)return 8; 106 if(x<=12&&y<=8)return 9; 107 if(x<=12&&y<=12)return 10; 108 if(x<=12)return 11; 109 110 if(y<=4)return 12; 111 if(y<=8)return 13; 112 if(y<=12)return 14; 113 return 15; 114 } 115 116 char str[260],s[17]; 117 int main() 118 { 119 while(true){ 120 int x=1,y=1; 121 DLX.Init(4096,1024); 122 for(int i=1;i<256;i+=16){ 123 if(not~scanf("%s",s))return 0; 124 for(int j=i;j<i+16;j++) 125 str[j]=s[j-i]; 126 } 127 for(int i=1;i<=256;i++) 128 { 129 for(int j=(i-1)*16+1;j<=i*16;j++) 130 { 131 int Color=j-(i-1)*16; 132 if(str[i]!='-'&&str[i]-'A'+1!=Color) 133 continue; 134 135 DLX.Link(j,(x-1)*16+Color); //行中对应颜色 136 DLX.Link(j,256+(y-1)*16+Color); //列中对应颜色 137 DLX.Link(j,512+Area(x,y)*16+Color);//块中对应颜色 138 DLX.Link(j,768+i); //矩阵中对应位置 139 } 140 y++;x+=y/17;y=(y-1)%16+1; 141 } 142 DLX.Solve(); 143 DLX.Print(); 144 } 145 return 0; 146 }