CSU 1969 TFSudoku 特殊数独

题目：

Description

At some point or another, most computer science students have written a standard Sudoku solving program. A slight twist has been added to standard Sudoku to make it a bit more challenging.

Digits from 1 to 9 are entered in a 6x6 grid so that no number is repeated in any row, column or 3x2 outlined region as shown below. Some squares in the grid are split by a slash and need 2 digits entered in them. The smaller number always goes above the slash.

For this problem, you will write a program that takes as input an incomplete puzzle grid and outputs the puzzle solution grid.

Input

The first line of input contains a single decimal integer P, (1 ≤ P ≤ 100), which is the number of data sets that follow. Each data set should be processed identically and independently.

Each data set consists of 7 lines of input. The first line of the data set contains the data set number, K. The remaining 6 lines represent an incomplete Tight-Fit Sudoku grid, each line has 6 data elements, separated by spaces. A data element can be a digit (1-9), a dash ('-') for a blank square or two of these separated by a slash ('/')

Output

For each data set there are 7 lines of output. The first output line consists of the data set number, K. The following 6 lines of output show the solution grid for the corresponding input data set. Each line will have 6 data elements, separated by spaces. A data element can be a digit (1-9), or 2 digits separated by a slash (‘/’).

Sample Input

11-/- -/5 4 3 2 -/-- 6 -/- -/- - -/-- 7/- - -/- -/- 28 -/- -/- - -/3 --/- - -/- -/- 4 --/- 8 7 6 5/- -/-

Sample Output

17/9 1/5 4 3 2 6/8 3 6 2/8 1/9 7 4/5 1 7/9 3 4/5 6/8 2 8 2/4 5/6 7 1/3 9 5/6 3 1/9 2/8 4 7 2/4 8 7 6 5/9 1/3 

一个加了特殊规则的数独，6x6的格子中会有斜杠出现，出现斜杠的格子斜杠上下一起要填2个数字，且上面的必须比下面小。

直接枚举不知道能不能过，我采用模拟人做数独的方法，先使用排除法，把能确定的数字都填上，没有可以确定的数字之后再选择可能性比较少的格子进行假设。

#include <iostream>#include <stdio.h>#include <string.h>#include <stack>#include <math.h>#include <algorithm>#include <map>#include <string>#define INF 0x3f3f3f3f#define MS(x,y) memset(x, y, sizeof(x))#define MOD 1000000007#define LL long long intusing namespace std;struct node{    int x,y;}z[10][10];int f[10][10][2][10];//前2维表示第几行第几列。第3维表示格子的上下。第4维表示那些数字不能填，0记录剩余可填的数字数bool xie[10][10];//记录有斜杠的格子int T;char s[5];void show(){    //printf("*****************************\n");    for(int i=0;i<6;++i)    {        for(int j=0;j<6;++j)        {            if(xie[i][j])            {                printf("%d/%d ", z[i][j].x, z[i][j].y);            }            else                printf("%d ", z[i][j].x);        }        printf("\n");    }    //printf("*****************************\n");}void tian(int a, int b, int x, int y)//填数字，并且把周围的格子出现这个数字的可能性去掉,x表示斜杠上的数字，y表示斜杠下的数字{    int xoff = b/3*3;    int yoff = a/2*2;    if(x != 0)    {        z[a][b].x = x;        for(int i=0;i<6;++i)        {            if(f[a][i][0][x] == 0)                --f[a][i][0][0];            f[a][i][0][x] = 1;            if(f[i][b][0][x] == 0)                --f[i][b][0][0];            f[i][b][0][x] = 1;            if(f[a][i][1][x] == 0)                --f[a][i][1][0];            f[a][i][1][x] = 1;            if(f[i][b][1][x] == 0)                --f[i][b][1][0];            f[i][b][1][x] = 1;        }        for(int i=0;i<2;++i)        {            for(int j=0;j<3;++j)            {                if(f[yoff+i][xoff+j][0][x] == 0)                    --f[yoff+i][xoff+j][0][0];                f[yoff+i][xoff+j][0][x] = 1;                if(f[yoff+i][xoff+j][1][x] == 0)                    --f[yoff+i][xoff+j][1][0];                f[yoff+i][xoff+j][1][x] = 1;            }        }        if(xie[a][b])        {            for(int i=x-1;i>0;--i)            {                if(f[a][b][1][i] == 0)                    --f[a][b][1][0];                f[a][b][1][i] = 1;            }        }    }    if(y != 0 && x != y)    {        z[a][b].y = y;        for(int i=0;i<6;++i)        {            if(f[a][i][0][y] == 0)                --f[a][i][0][0];            f[a][i][0][y] = 1;            if(f[i][b][0][y] == 0)                --f[i][b][0][0];            f[i][b][0][y] = 1;            if(f[a][i][1][y] == 0)                --f[a][i][1][0];            f[a][i][1][y] = 1;            if(f[i][b][1][y] == 0)                --f[i][b][1][0];            f[i][b][1][y] = 1;        }        for(int i=0;i<2;++i)        {            for(int j=0;j<3;++j)            {                if(f[yoff+i][xoff+j][0][y] == 0)                    --f[yoff+i][xoff+j][0][0];                f[yoff+i][xoff+j][0][y] = 1;                if(f[yoff+i][xoff+j][1][y] == 0)                    --f[yoff+i][xoff+j][1][0];                f[yoff+i][xoff+j][1][y] = 1;            }        }        if(xie[a][b])        {            for(int i=y+1;i<10;++i)            {                if(f[a][b][0][i] == 0)                    --f[a][b][0][0];                f[a][b][0][i] = 1;            }        }    }}bool judge()//判断是否得出答案{    for(int i=0;i<6;++i)    {        for(int j=0;j<6;++j)        {            if(xie[i][j])            {                if(z[i][j].x == 0 || z[i][j].y == 0)                    return false;            }            else            {                if(z[i][j].x == 0)                    return false;            }        }    }    return true;}bool solve(){    while(1)    {        int x=0,y=0,MIN = 9,ff = 0;        bool istian = false;        bool maodun = false;        for(int i=0;i<6;++i)        {            for(int j=0;j<6;++j)            {                if(xie[i][j])                {                    if(z[i][j].x == 0)                    {                        if(f[i][j][0][0] == 0)                        {                            maodun = true;                            //printf("*****(%d,%d)矛盾\n", i, j);                            break;                        }                        else if(f[i][j][0][0] == 1)                        {                            for(int ii=1;ii<10;++ii)                            {                                if(f[i][j][0][ii] == 0)                                {                                    tian(i,j,ii,0);                                    istian = true;                                }                            }                        }                        else if(f[i][j][0][0] > 1)                        {                            if(f[i][j][0][0] < MIN)                            {                                MIN = f[i][j][0][0];                                x = j;                                y = i;                                ff = 0;                            }                        }                    }                    if(z[i][j].y == 0)                    {                        if(f[i][j][1][0] == 0)                        {                            maodun = true;                            //printf("*****(%d,%d)矛盾\n", i, j);                            break;                        }                        else if(f[i][j][1][0] == 1)                        {                            for(int ii=1;ii<10;++ii)                            {                                if(f[i][j][1][ii] == 0)                                {                                    tian(i,j,0,ii);                                    istian = true;                                }                            }                        }                        else if(f[i][j][1][0] > 1)                        {                            if(f[i][j][1][0] < MIN)                            {                                MIN = f[i][j][1][0];                                x = j;                                y = i;                                ff = 1;                            }                        }                    }                }                else                {                    if(z[i][j].x == 0)                    {                        if(f[i][j][0][0] == 0)                        {                            maodun = true;                            //printf("*****(%d,%d)矛盾\n", i, j);                            break;                        }                        else if(f[i][j][0][0] == 1)                        {                            for(int ii=1;ii<10;++ii)                            {                                if(f[i][j][0][ii] == 0)                                {                                    tian(i,j,ii,ii);                                    istian = true;                                }                            }                        }                        else if(f[i][j][0][0] > 1)                        {                            if(f[i][j][0][0] < MIN)                            {                                MIN = f[i][j][0][0];                                x = j;                                y = i;                                ff = 0;                            }                        }                    }                }            }            if(maodun)                break;        }        //show();        if(maodun)        {            //printf("返回\n");            return false;        }        if(!istian)        {            if(judge()) return true;            node tz[10][10];            int tf[10][10][2][10];            for(int i=0;i<6;++i)            {                for(int j=0;j<6;++j)                {                    tz[i][j] = z[i][j];                    for(int k=0;k<2;++k)                    {                        for(int m=0;m<10;++m)                        {                            tf[i][j][k][m] = f[i][j][k][m];                        }                    }                }            }            for(int ii=1;ii<10;++ii)            {                if(f[y][x][ff][ii] == 0)                {                    //printf("*************假设(%d,%d,%d,%d)\n", y, x, ff, ii);                    if(y == 3 && x == 1 && ff == 0 && ii == 2)                        ii = 2;                    if(ff == 0) tian(y,x,ii,0);                    else tian(y,x,0,ii);                    if(solve())                        return true;                    for(int i=0;i<6;++i)                    {                        for(int j=0;j<6;++j)                        {                            z[i][j] = tz[i][j];                            for(int k=0;k<2;++k)                            {                                for(int m=0;m<10;++m)                                {                                    f[i][j][k][m] = tf[i][j][k][m];                                }                            }                        }                    }                }            }            return false;        }    }    return true;}int main(){    scanf("%d", &T);    while(T--)    {        int CASE;        scanf("%d", &CASE);        MS(z, 0);        MS(f, 0);        MS(xie, 0);        for(int i=0;i<6;++i)        {            for(int j=0;j<6;++j)            {                f[i][j][0][0] = f[i][j][1][0] = 9;            }        }        for(int i=0;i<6;++i)        {            for(int j=0;j<6;++j)            {                scanf("%s", s);                int len = strlen(s);                if(len == 1 && s[0] != '-')                {                    tian(i,j,s[0]-'0',s[0]-'0');                }                if(len == 3)                {                    xie[i][j] = 1;                    tian(i,j,s[0]=='-'?0:s[0]-'0',s[2]=='-'?0:s[2]-'0');                }            }            int t = 0;        }        //show();        solve();        printf("%d\n", CASE);        show();    }    return 0;}