From 2b7444ab92f90350d24279ec556bcafeb395dd1b Mon Sep 17 00:00:00 2001 From: pm6lnt93v <1289474397@qq.com> Date: Tue, 5 Nov 2024 17:20:39 +0800 Subject: [PATCH] ADD file via upload --- 4.c | 255 ++++++++++++++++++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 255 insertions(+) create mode 100644 4.c diff --git a/4.c b/4.c new file mode 100644 index 0000000..bc26e66 --- /dev/null +++ b/4.c @@ -0,0 +1,255 @@ +#include + +int main() +{ + //引用函数 + void Printf(char* p0); + int Sudoku(char* p0); + void Sudokuprint(char* p0); + void Sudokuback(int k,char* p0); + //给出矩阵 并打印 + char board[9][9]={ + {'3','.','4','.','.','.','1','.','2'}, + {'.','.','.','2','3','8','.','.','.'}, + {'9','.','.','.','.','4','.','.','6'}, + {'.','3','5','.','7','.','.','6','.'}, + {'.','4','.','3','.','5','.','8','.'}, + {'.','7','.','.','9','.','2','5','.'}, + {'8','.','.','6','.','.','.','.','7'}, + {'.','.','.','9','4','7','.','.','.'}, + {'4','.','7','.','.','.','5','.','9'}}; + char* p0 = board; + printf("The original Sudoku matrix:\n"); + Printf(p0); + int i,j,k; + //成立的情况 运用回溯函数 + if(Sudoku(p0)) + { + printf("True:Valid initial Sudoku matrix!\n"); + Sudokuback(0, p0); + } + //不成立的情况 + else + { + printf("False:Invalid initial Sudoku matrix!\n"); + Sudokuprint(p0); + printf("No solution\n"); + } + return 0; +} +//打印矩阵 +void Printf(char* p0) +{ + int i,j; + for(i=0;i<9;i++) + { + for(j=0;j<9;j++) + { + printf("%c ",p0[9*i+j]); + } + printf("\n"); + } +} +//判断是数独但不打印 +int Sudoku(char* p0) +{ +int i,j,number,s,q,p,t,g; +//此处引入1 +int c[10]={0,0,0,0,0,0,0,0,0,0};//用来储存数字出现的个数 + + +//下面小矩阵的判断 +int w[10]={1,2,3,4,5,6,7,8,9}; +int k=0; +for(t=3;t<=9;t=t+3) +{ + for(g=3;g<=9;g=g+3) + { + for(i=t-3;i1) + { + return 0; + } + } + k++; + for(s=0;s<=9;s++)c[s]=0; + } + +} +//完成行的判断 +for(i=0;i<9;i++) +{ + + for(number=1;number<=9;number++) + { + for(j=0;j<9;j++) + { + if(p0[9*i+j] == number + '0') + c[number]++; + } + } + for(q=1;q<=9;q++) + { + if(c[q]>1) + { + return 0 ; + } +} + for(p=0;p<10;p++)c[p]=0; +} +//同理列的判断 +for(i=0;i<9;i++) +{ + for(number=1;number<=9;number++) + { + for(j=0;j<9;j++) + { + if(p0[9*j+i] == number + '0') + c[number]++;} + } + for(q=1;q<=9;q++) + { + if(c[q]>1) + { + return 0; + } + } + for(p=0;p<10;p++)c[p]=0; +} +return 1; +} +//判断数独并打印 +void Sudokuprint(char* p0) +{ + + int i, j, number, s, q, p, t, g; + //此处引入1 + int c[10] = { 0,0,0,0,0,0,0,0,0,0 };//用来储存数字出现的个数 + + + //下面小矩阵的判断 + int w[10] = { 1,2,3,4,5,6,7,8,9 }; + int k = 0; + for (t = 3;t <= 9;t = t + 3) + { + for (g = 3;g <= 9;g = g + 3) + { + for (i = t - 3;i < t;i++) + { + for (j = g - 3;j < g;j++)//分出来小矩阵 + { + for (number = 1;number <= 9;number++) + { + if (p0[9 * i + j] == number + '0') + { + c[number]++; + } + } + } + } + for (q = 1;q <= 9;q++) + { + if (c[q] > 1) + { + printf("The number %d in the block %d has been used!\n", q, w[k]); + return ; + } + } + k++; + for (s = 0;s <= 9;s++)c[s] = 0; + } + + } + //完成行的判断 + for (i = 0;i < 9;i++) + { + + for (number = 1;number <= 9;number++) + { + for (j = 0;j < 9;j++) + { + if (p0[9 * i + j] == number + '0') + c[number]++; + } + } + for (q = 1;q <= 9;q++) + { + if (c[q] > 1) + { + printf("The number %d in the col %d has been used!\n", q, i + 1); + return ; + } + } + for (p = 0;p < 10;p++)c[p] = 0; + } + //同理列的判断 + for (i = 0;i < 9;i++) + { + for (number = 1;number <= 9;number++) + { + for (j = 0;j < 9;j++) + { + if (p0[9 * j + i] == number + '0') + c[number]++; + } + } + for (q = 1;q <= 9;q++) + { + if (c[q] > 1) + { + printf("The number %d in the column %d has been used!\n", q, i + 1); + return ; + } + } + for (p = 0;p < 10;p++)c[p] = 0; + } + return ; +} +//回溯法填空 +void Sudokuback(int k,char* p0) +{ + int m; + if (k == 81) + { + if (Sudoku(p0)) + { + printf("The solution of Sudoku matrix:\n"); + Printf(p0); + } + else printf("No sulotion\n"); + return; + } + if (p0[k] == '.') + { + for (m = 1;m <= 9;m++) + { + p0[k] = m+'0'; + if (Sudoku(p0)) + { + Sudokuback(k + 1,p0); + } + + else if (!Sudoku(p0)&&m==9) + { + p0[k] = '.'; + return ; + } + } + } + else + { + Sudokuback(k + 1, p0); + } +} \ No newline at end of file