回朔算法2

{
//獲取題目信息
int data[]={0,5,0,0,3,0,0,0,0,
0,0,1,6,0,0,9,0,8,
3,0,0,0,0,0,0,0,0,
0,0,0,8,0,0,0,0,0,
0,0,0,0,0,0,7,0,0,
0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0};
int i,j,z=0;
for(i=0;i<9;i++)
for(j=0;j<9;j++)
{
SD[i][j]=data[z];
z++;
}
sd_in();

}

void TuiLi::sd_out()
{
int i,j,z=0;
for(i=0;i<9;i++)
for(j=0;j<9;j++)
{
SD[i][j]=sudu[z];
z++;
}
}

void TuiLi::sd_print()
{
int i,j,z=0;
sd_out();
GetLocalTime(&TIME);//獲得系統當前時(shí)間
printf("找到一個(gè)數獨解獲取時(shí)間為：%d年%d月%d日%d時(shí)%d分%d秒",TIME.wYear,TIME.wMonth,TIME.wDay,TIME.wHour,TIME.wMinute,TIME.wSecond);
for(i=0;i<9;i++)
{
printf(" ");
for(j=0;j<9;j++)
{
printf("%d ",SD[i][j]);

}
printf("");
}
printf("");
}

int TuiLi::ok_sd() //解數獨用剪枝函數
{
int i,j,x;
sd_out();

for(x=0;x<9;x++)
for(i=0;i<9;i++)
for(j=i+1;j<9;j++)
{
if(SD[x][i]==SD[x][j]&&SD[x][i]!=0) return 1;
}
//不允許行重復

for(x=0;x<9;x++)
for(i=0;i<9;i++)
for(j=i+1;j<9;j++)
{
if(SD[i][x]==SD[j][x]&&SD[i][x]!=0) return 1;
}
//不允許列重復

return 0;
}

int TuiLi::go_sd() //解數獨入口函數
{
int i,c=0;
//獲取題目信息
SD_in();
//回溯根節點(diǎn)初始值，調整這個(gè)值能調整回溯進(jìn)度
if(!sudu_ji[0])sudu[0]=1;
i=0;
while(1)
{
//sd_print();
//Sleep(10);
if(sudu[i]<=9)
{
//回溯點(diǎn)取值在正常范圍內
if(ok_sd())
{
//如果和剪枝條件沖突，嘗試下一個(gè)可用值
if(!sudu_ji[i])sudu[i]++;
continue;
}
if(i>=80)
{
//已經(jīng)到了最后一個(gè)節點(diǎn)也就是找到了一個(gè)解先輸出它
sd_print();
//然后嘗試找到其它的解
if(!sudu_ji[80])sudu[80]++;
c++;
if(c>=10)
{
printf("此題解太多只輸出前面10個(gè)解!");
return 0;
}
continue;
}
//回溯點(diǎn)值沒(méi)有產(chǎn)生沖突擴展他嘗試開(kāi)始下一個(gè)節點(diǎn)
i++;
if(!sudu_ji[i])sudu[i]=1;
}else
{
//超過(guò)回溯點(diǎn)取值范圍
//此回溯點(diǎn)已死復位
if(!sudu_ji[i])sudu[i]=0;
//退回前一回溯點(diǎn)
i--;
if(sudu_ji[i]) i--;//回溯后的當前回溯節點(diǎn)是題目已經(jīng)標記過(guò)的就在往前回溯一步

if(i<0)
{
printf("已經(jīng)找出數獨所有解共有：%d個(gè)解！",c);
return 1;
}
//嘗試對上一回溯點(diǎn)進(jìn)行擴展
if(!sudu_ji[i])sudu[i]++;
continue;
}
}

return 0;
}

void TuiLi::hh_init()
{
//數據初始化
int i,j;
printf("請輸入皇后問(wèn)題的解題規模：");
scanf("%d",&N);
printf("");
for(i=0;i<1024;i++)
{
hh[i]=0;
}
for(i=0;i<30;i++)
for(j=0;j<30;j++)
{
HH[i][j]=0;
}
}