90 分#include
     
      #include
      
       using namespace std;int flag[99][99];int s=0;int a[9][9]={ {6,6,6,6, 6,6,6,6,6},{6,7,7,7, 7,7,7,7,6},{6,7,8,8, 8,8,8,7,6},{6,7,8,9, 9,9,8,7,6},{6,7,8,9,10,9,8,7,6},{6,7,8,9, 9,9,8,7,6},{6,7,8,8, 8,8,8,7,6},{6,7,7,7, 7,7,7,7,6},{6,6,6,6, 6,6,6,6,6}};int c[9][9]={    1,1,1,2,2,2,3,3,3,    1,1,1,2,2,2,3,3,3,    1,1,1,2,2,2,3,3,3,    4,4,4,5,5,5,6,6,6,    4,4,4,5,5,5,6,6,6,    4,4,4,5,5,5,6,6,6,    7,7,7,8,8,8,9,9,9,    7,7,7,8,8,8,9,9,9,    7,7,7,8,8,8,9,9,9,};int b[99][99];int f[10][10];int fh[10][10];int fs[10][10];int max1=-1;int f4;void dfs(int x,int y,int ans){    if(f4==1) return;        if(x==0&&y==9)    {   f4=1;        for(int i=0;i<=8;i++)        {
                 for(int j=0;j<=8;j++)             printf("%d ",b[i][j]);            printf("\n");        }        //max1=max(ans,max1);       // return;    }        if(!b[x][y])     {           for(int i=1;i<=9;i++)        if(fh[x][i]==0&&fs[y][i]==0&&f[c[x][y]][i]==0)        {
                 b[x][y]=i;                fh[x][i]=1;                fs[y][i]=1;                f[c[x][y]][i]=1;        if(x<8)            {
                     dfs(x+1,y,ans+a[x][y]*i);            }            else {
                     dfs(0,y+1,ans+a[x][y]*i);            }                fh[x][i]=0;                fs[y][i]=0;                f[c[x][y]][i]=0;                b[x][y]=0;        }        }        else        {
                 if(x<8)            dfs(x+1,y,ans);            else dfs(0,y+1,ans);        }}int main(){    for(int i=0;i<=8;i++)    for(int j=0;j<=8;j++)     {
             scanf("%d",&b[i][j]);        if(b[i][j])        {
             fh[i][b[i][j]]=1;        fs[j][b[i][j]]=1;        f[c[i][j]][b[i][j]]=1;        s+=a[i][j]*b[i][j];        }    }    dfs(0,0,s);    printf("%d",max1);    return 0;}
      
     

可以解数独游戏了！！！

这个题一开始我是蛇形遍历，有点麻烦

。。。

后来看了xy的便改了

sxb大神思路#include
     
      #include
      
       #include
       
        using namespace std;int s=0;int n=0;int can[10][10];//记录能填多少数 int a[9][9]={ {6,6,6,6, 6,6,6,6,6},{6,7,7,7, 7,7,7,7,6},{6,7,8,8, 8,8,8,7,6},{6,7,8,9, 9,9,8,7,6},{6,7,8,9,10,9,8,7,6},{6,7,8,9, 9,9,8,7,6},{6,7,8,8, 8,8,8,7,6},{6,7,7,7, 7,7,7,7,6},{6,6,6,6, 6,6,6,6,6}};//打表 int c[9][9]={    1,1,1,2,2,2,3,3,3,    1,1,1,2,2,2,3,3,3,    1,1,1,2,2,2,3,3,3,    4,4,4,5,5,5,6,6,6,    4,4,4,5,5,5,6,6,6,    4,4,4,5,5,5,6,6,6,    7,7,7,8,8,8,9,9,9,    7,7,7,8,8,8,9,9,9,    7,7,7,8,8,8,9,9,9,};int b[99][99];int f[10][10];int fh[10][10];int fs[10][10];int max1=-1;int f1(int x,int y,int k){    if(!fh[x][k]&&!fs[y][k]&&!f[c[x][y]][k])        return 1;    return 0;}void dfs(int X,int ans){        if(X>(81-n))//已填完     {
             max1=max(ans,max1);        return;    }     memset(can,0,sizeof(can));     int x,y,mmin=99999999;    for(int i=0;i<9;i++)    for(int j=0;j<9;j++)    if(!b[i][j])    {
             for(int k=1;k<=9;k++)        if ( f1(i,j,k) ) can[i][j]++;        if(can[i][j]
       
      
     

这个思路666，直接就a了otc