皇后问题(源程序).doc

1、#include #include #include #define StackSize 20typedef struct int row,col;DataType;typedef struct DataType dataStackSize;int top;SeqStack;typedef structint chess2020;QueenType;void main()SeqStack S;DataType newelem;QueenType Q;int row,col,number;void InitStack(SeqStack *S, QueenType *Q, int number);

2、void Push(SeqStack *S, DataType newelem, QueenType *Q, int number);DataType Pop(SeqStack *S, QueenType *Q, int number);void PrintStack(SeqStack S);docoutnumber;if (number!=0)InitStack(row=0;col=0;while (row!=-1)while (col0)col+;if (col=number) /回溯到上一个皇后newelem=Pop(row-;col=newelem.col+1;else /放置皇后ne

3、welem.row=row;newelem.col=col;Push(row+;col=0;if (row=number) /皇后已放满，输出，退后二步PrintStack(S);newelem=Pop( /最后一个皇后newelem=Pop( /最后第二个皇后row=newelem.row;col=newelem.col+1;while(number!=0);void InitStack(SeqStack *S, QueenType *Q, int number)int i,j;S-top=-1;for (i=0;ichessij=0;void Push(SeqStack *S, DataT

4、ype newelem, QueenType *Q, int number) /放置皇后时，在四条线上各单元加 1， （行、列、左上右下、右上左下）int i,j,k;S-top+;S-dataS-top=newelem;for (i=0;ichessinewelem.col+;for (j=0;jchessnewelem.rowj+;if (newelem.rowchessij+;i+;j+;if (newelem.row=0)Q-chessij+;i+;j-;DataType Pop(SeqStack *S, QueenType *Q, int number) /撤销皇后时，在四条线上各单

5、元减 1， （行、列、左上右下、右上左下）int i,j,k;DataType elem;S-top-;elem=S-dataS-top+1;for (i=0;ichessielem.col-;for (j=0;jchesselem.rowj-;if (elem.rowchessij-;i+;j+;if (elem.row=0)Q-chessij-;i+;j-;return elem;void PrintStack(SeqStack S)int i;cout“ 答案为：“;for (i=0; i=S.top;i+)cout“(“S.datai.row“,“S.datai.col“) “;coutendl;