数独问题

//数独 #include<stdio.h>#define n 9int a[n][n] = {0,0,5,3,0,0,0,0,0,8,0,0,0,0,0,0,2,0,0,7,0,0,1,0,5,0,0,4,0,0,0,0,5,3,0,0,0,1,0,0,7,0,0,0,6,0,0,3,2,0,0,0,8,0,0,6,0,5,0,0,0,0,9,0,0,4,0,0,0,0,3,0,0,0,0,0,0,9,7,0,0}; bool ok(int x,int y){int x0,y0;//m用来表示当前（x,y）所在的小的九宫格的左上角的坐标int i,j;//对行进行判断 for(i = 0;i < n;i ++){if(y != i && a[x][i] == a[x][y]){return false;}}//对列进行判断 for(i = 0;i < n;i ++){if(x != i && a[i][y] == a[x][y]){return false;}}x0 = x / 3 * 3;y0 = y / 3 * 3;//对小九宫格进行判断 for(i = x0;i < x0 + 3;i ++){for(j = y0;j < y0 + 3;j ++){if(a[i][j] == a[x][y] && x != i && y != j){return false;}}}return true;}void nfs(int t){int i,j;int x,y;if(t == n * n){for(i = 0;i < n;i ++){for(j = 0;j < n;j ++){printf("%d\t",a[i][j]);}putchar('\n');}return;}x = t / n;y = t % n;if(a[x][y] == 0){//应该先进行判断当前位置是否存在数，在不存在的时候再进行往下分支 for(i = 1;i <= n;i ++){a[x][y] = i;if(ok(x,y)){nfs(t + 1);}a[x][y] = 0;}}else{nfs(t + 1);}}int main(){nfs(0);return 0;}===========================================================================================//数独，以坐标为t的回溯 #include<stdio.h>#define n 9int x0 = 0,y0 = 0,x1,y1;int a[n][n] = {0,0,5,3,0,0,0,0,0,8,0,0,0,0,0,0,2,0,0,7,0,0,1,0,5,0,0,4,0,0,0,0,5,3,0,0,0,1,0,0,7,0,0,0,6,0,0,3,2,0,0,0,8,0,0,6,0,5,0,0,0,0,9,0,0,4,0,0,0,0,3,0,0,0,0,0,0,9,7,0,0}; int flag = 0;bool ok(int x,int y){int x0,y0;//m用来表示当前（x,y）所在的小的九宫格的右上角的坐标int i,j;//对行进行判断 for(i = 0;i < n;i ++){if(y != i && a[x][i] == a[x][y]){return false;}}//对列进行判断 for(i = 0;i < n;i ++){if(x != i && a[i][y] == a[x][y]){return false;}}x0 = x / 3 * 3;y0 = y / 3 * 3;//对小九宫格进行判断 for(i = x0;i < x0 + 3;i ++){for(j = y0;j < y0 + 3;j ++){if(a[i][j] == a[x][y] && x != i && y != j){return false;}}}return true;}void nfs(int x,int y){//以列优先进行判断 int i,j;if(flag)return;if(a[x][y] != 0){//当前位置放数了 if(y == n - 1){//最后一列 if(x == n - 1){//最后一行 flag = 1;for(i = 0;i < n;i ++){for(j = 0;j < n;j ++){printf("%d\t",a[i][j]);}putchar('\n');}return;}else{//到最后一列了，但是没有到最后一行 nfs(x + 1,y);}}else{//没有到最后一列 if(x == n - 1){//到最后一行 nfs(0,y + 1);}else{//没有到最后一行 nfs(x + 1,y);}} }else{// 当前位置没有放数if(y == n - 1){//最后一列 if(x == n - 1){//最后一行 for(i = 1;i <= n;i ++){a[x][y] = i;if(ok(x,y)){//判断放进去的数是否符合条件 flag = 1;for(i = 0;i < n;i ++){for(j = 0;j < n;j ++){printf("%d\t",a[i][j]);}putchar('\n');}return;}a[x][y] = 0;}}else{//没有到最后一行 for(i = 1;i <= n;i ++){a[x][y] = i;if(ok(x,y)){nfs(x + 1,y);}a[x][y] = 0;}}}else{//不是最后一列 if(x == n - 1){//最后一行 for(i = 1;i <= n;i ++){a[x][y] = i;if(ok(x,y)){nfs(0,y + 1);}a[x][y] = 0;}}else{//不是最后一行 for(i = 1;i <= n;i ++){a[x][y] = i;if(ok(x,y)){nfs(x + 1,y);}a[x][y] = 0;}}}}}int main(){nfs(x0,y0);return 0;}