求教：写八皇后时遇到的奇葩问题！ - C语言论坛

楼主

问题点数：0 回复次数：5

求教：写八皇后时遇到的奇葩问题！

第一次我让整形变量num作为计数器，结果让我费解；第二次我让answer[0].n作为计数器，正常。

程序代码：

#include<stdio.h>
#define N 8
typedef struct
{
    int n;
    int row;
    int line;
}AnswerType;
AnswerType answer[N];
typedef enum
{
    notoccued = 0,
    occued = 1
}Ifoccued;
Ifoccued RowOccued[N];
Ifoccued LeftTop_RightDown[2*N-1];
Ifoccued LeftDown_RightTop[2*N-1];
void print();
int num;
void start(int LineIndex)
{
    for(int RowIndex=0;RowIndex<8;RowIndex++)
    {
        if(RowOccued[RowIndex]==notoccued&&LeftTop_RightDown[RowIndex-LineIndex+N-1]==notoccued
            &&LeftDown_RightTop[RowIndex+LineIndex]==notoccued)
        {
            RowOccued[RowIndex]=occued;
            LeftTop_RightDown[RowIndex-LineIndex+N-1]=occued;
            LeftDown_RightTop[RowIndex+LineIndex]=occued;
            answer[LineIndex].row=RowIndex;
            answer[LineIndex].line=LineIndex;
            if(LineIndex==7)
            { 
                num++;
                answer[0].n++;
                print();
                RowOccued[N]=notoccued;
                LeftTop_RightDown[2*N-1]=notoccued;
                LeftDown_RightTop[2*N-1]=notoccued;
            }else
            {
                start(LineIndex+1);
            }
        RowOccued[RowIndex]=notoccued;
        LeftTop_RightDown[RowIndex-LineIndex+N-1]=notoccued;
        LeftDown_RightTop[RowIndex+LineIndex]=notoccued;
        }
    }
}
void print()
{
    printf("第%d个结果为：",answer[0].n);
    for(int i=0;i<N;i++)
    {
        printf("[%d,%d] ",answer[i].line+1,answer[i].row+1);
    }
    printf("\n");
}
int main()
{
    num=0;
    answer[0].n=0;
    start(0);
    return 0;
}

这是上面代码的运行截图，没有问题 92个结果
[local]1[/local]

注意下面我把计数器换成了num

程序代码：

#include<stdio.h>
#define N 8
typedef struct
{
    int n;
    int row;
    int line;
}AnswerType;
AnswerType answer[N];
typedef enum
{
    notoccued = 0,
    occued = 1
}Ifoccued;
Ifoccued RowOccued[N];
Ifoccued LeftTop_RightDown[2*N-1];
Ifoccued LeftDown_RightTop[2*N-1];
void print();
int num;
void start(int LineIndex)
{
    for(int RowIndex=0;RowIndex<8;RowIndex++)
    {
        if(RowOccued[RowIndex]==notoccued&&LeftTop_RightDown[RowIndex-LineIndex+N-1]==notoccued
            &&LeftDown_RightTop[RowIndex+LineIndex]==notoccued)
        {
            RowOccued[RowIndex]=occued;
            LeftTop_RightDown[RowIndex-LineIndex+N-1]=occued;
            LeftDown_RightTop[RowIndex+LineIndex]=occued;
            answer[LineIndex].row=RowIndex;
            answer[LineIndex].line=LineIndex;
            if(LineIndex==7)
            { 
                num++;
                answer[0].n++;
                print();
                RowOccued[N]=notoccued;
                LeftTop_RightDown[2*N-1]=notoccued;
                LeftDown_RightTop[2*N-1]=notoccued;
            }else
            {
                start(LineIndex+1);
            }
        RowOccued[RowIndex]=notoccued;
        LeftTop_RightDown[RowIndex-LineIndex+N-1]=notoccued;
        LeftDown_RightTop[RowIndex+LineIndex]=notoccued;
        }
    }
}
void print()
{
    printf("第%d个结果为：",num);
    for(int i=0;i<N;i++)
    {
        printf("[%d,%d] ",answer[i].line+1,answer[i].row+1);
    }
    printf("\n");
}
int main()
{
    num=0;
    answer[0].n=0;
    start(0);
    return 0;
}

结果竟然是这样

搜索更多相关主题的帖子: 计数器　皇后　 color　

第 2 楼

得分:0

结果没区别啊，楼主

[fly]存在即是合理[/fly]

第 3 楼

得分:0

说了 num 要初始化

第 4 楼

得分:0

我想起了最难读懂的小程序就是八皇后问题

我了个去的编程！

第 5 楼

得分:0

void print()
{
    printf("第%d个结果为：",num);
    for(int i=0;i<N;i++)
    {
        printf("[%d,%d] ",answer[i].line+1,answer[i].row+1);
    }
    printf("\n");
}
------
这个Num哪来的，
找你的代码是void print();
int num;这个吧
全局变量。。。。。。。。。。

第 6 楼

得分:0

#include<stdio.h>
#define N 8
int num = 0;
void print();

typedef struct
{
    int n;
    int row;
    int line;
}AnswerType;
AnswerType answer[N];
typedef enum
{
    notoccued = 0,
    occued = 1
}Ifoccued;

Ifoccued RowOccued[N];
Ifoccued LeftTop_RightDown[2*N-1];
Ifoccued LeftDown_RightTop[2*N-1];

void start(int LineIndex)
{
    for(int RowIndex=0;RowIndex<8;RowIndex++)
    {
        if(RowOccued[RowIndex]==notoccued&&LeftTop_RightDown[RowIndex-LineIndex+N-1]==notoccued
            &&LeftDown_RightTop[RowIndex+LineIndex]==notoccued)
        {
            RowOccued[RowIndex]=occued;
            LeftTop_RightDown[RowIndex-LineIndex+N-1]=occued;
            LeftDown_RightTop[RowIndex+LineIndex]=occued;
            answer[LineIndex].row=RowIndex;
            answer[LineIndex].line=LineIndex;
            if(LineIndex==7)
            {
                num++;
                answer[0].n++;
                print();
                RowOccued[N]=notoccued;
                LeftTop_RightDown[2*N-1]=notoccued;
                LeftDown_RightTop[2*N-1]=notoccued;
            }else
            {
                start(LineIndex+1);
            }
        RowOccued[RowIndex]=notoccued;
        LeftTop_RightDown[RowIndex-LineIndex+N-1]=notoccued;
        LeftDown_RightTop[RowIndex+LineIndex]=notoccued;
        }
    }
}

void print()
{
    int a = 1;
   printf("第%d个结果为：\n",a++);
    for(int i=0;i<N;i++)
    {
        printf("[%d,%d] ",answer[i].line+1,answer[i].row+1);
    }
    printf("\n");
}

int main()
{

    num=0;
    answer[0].n=0;
    start(0);
    return 0;
}

---------
int sum = 0；