2012年第三届蓝桥杯C/C++程序设计本科B组省赛 放棋子(代码填空)

2012年第三届蓝桥杯C/C++程序设计本科B组省赛题目汇总：

http://blog.csdn.net/u014552756/article/details/50583827

放棋子    
今有 6 x 6 的棋盘格。其中某些格子已经预先放好了棋子。现在要再放上去一些，使得：每行每列都正好有3颗棋子。我们希望推算出所有可能的放法。下面的代码就实现了这个功能。

初始数组中，“1”表示放有棋子，“0”表示空白。

int N = 0;bool CheckStoneNum(int x[][6]){for(int k=0; k<6; k++){int NumRow = 0;int NumCol = 0;for(int i=0; i<6; i++){if(x[k][i]) NumRow++;if(x[i][k]) NumCol++;}if(_____________________) return false;  // 填空}return true;}int GetRowStoneNum(int x[][6], int r){int sum = 0;for(int i=0; i<6; i++) if(x[r][i]) sum++;return sum;}int GetColStoneNum(int x[][6], int c){int sum = 0;for(int i=0; i<6; i++) if(x[i][c]) sum++;return sum;}void show(int x[][6]){for(int i=0; i<6; i++){for(int j=0; j<6; j++) printf("%2d", x[i][j]);printf("\n");}printf("\n");}void f(int x[][6], int r, int c);void GoNext(int x[][6],  int r,  int c){if(c<6)_______________________;   // 填空elsef(x, r+1, 0);}void f(int x[][6], int r, int c){if(r==6){if(CheckStoneNum(x)){N++;show(x);}return;}if(______________)  // 已经放有了棋子{GoNext(x,r,c);return;}int rr = GetRowStoneNum(x,r);int cc = GetColStoneNum(x,c);if(cc>=3)  // 本列已满GoNext(x,r,c);  else if(rr>=3)  // 本行已满f(x, r+1, 0);   else{x[r][c] = 1;GoNex。t(x,r,c);x[r][c] = 0;if(!(3-rr >= 6-c || 3-cc >= 6-r))  // 本行或本列严重缺子，则本格不能空着！GoNext(x,r,c);  }}int main(int argc, char* argv[]){int x[6][6] = {{1,0,0,0,0,0},{0,0,1,0,1,0},{0,0,1,1,0,1},{0,1,0,0,1,0},{0,0,0,1,0,0},{1,0,1,0,0,1}};f(x, 0, 0);printf("%d\n", N);return 0;}<span style="font-family:Microsoft YaHei;"></span>

思路：
每个函数的作用：
bool CheckStoneNum(int x[][6]);                  判断每行每列是否都正好有3颗棋子；
int GetRowStoneNum(int x[][6], int r);          返回r行存在的棋子数；
int GetColStoneNum(int x[][6], int c);           返回c列的棋子数；
void show(int x[][6]);                                        将符合要求的方案打印出来；
void GoNext(int x[][6],  int r,  int c);                 从(r, c)开始往后遍历二维数组；

答案：NumRow != 3 || NumCol != 3
           f(x, r, c + 1)
           x[r][c] == 1

# include <stdio.h>int N = 0;bool CheckStoneNum(int x[][6]){for(int k=0; k<6; k++){int NumRow = 0;int NumCol = 0;for(int i=0; i<6; i++){if(x[k][i]) NumRow++;if(x[i][k]) NumCol++;}if(NumRow != 3 || NumCol != 3) return false;  // 填空}return true;}int GetRowStoneNum(int x[][6], int r){int sum = 0;for(int i=0; i<6; i++) if(x[r][i]) sum++;return sum;}int GetColStoneNum(int x[][6], int c){int sum = 0;for(int i=0; i<6; i++) if(x[i][c]) sum++;return sum;}void show(int x[][6]){for(int i=0; i<6; i++){for(int j=0; j<6; j++) printf("%2d", x[i][j]);printf("\n");}printf("\n");}void f(int x[][6], int r, int c);void GoNext(int x[][6],  int r,  int c){if(c<6)f(x, r, c + 1);   // 填空elsef(x, r+1, 0);}void f(int x[][6], int r, int c){if(r==6){if(CheckStoneNum(x)){N++;show(x);}return;}if(x[r][c] == 1)  // 已经放有了棋子{GoNext(x,r,c);return;}int rr = GetRowStoneNum(x,r);int cc = GetColStoneNum(x,c);if(cc>=3)  // 本列已满GoNext(x,r,c);  else if(rr>=3)  // 本行已满f(x, r+1, 0);   else{x[r][c] = 1;GoNext(x,r,c);x[r][c] = 0;if(!(3-rr >= 6-c || 3-cc >= 6-r))  // 本行或本列严重缺子，则本格不能空着！GoNext(x,r,c);  }}int main(int argc, char* argv[]){int x[6][6] = {{1,0,0,0,0,0},{0,0,1,0,1,0},{0,0,1,1,0,1},{0,1,0,0,1,0},{0,0,0,1,0,0},{1,0,1,0,0,1}};f(x, 0, 0);printf("%d\n", N);return 0;}