• NOIP2009 靶型数独


    传送门

    这道题比我想象之中要暴力一些。

    自己一开始有一份写9*9数独的代码,自己试了一下直接交上去只有40分。看来这样是肯定不行的。考虑优化,想在中间贪贪心啥的,但是难以保证正确性。最后学了一招,从数字比较多的行开始搜索,这样会使搜索树变得小一些,时间就会减少一些。

    然后交上去之后还是只有75分……发现自己数独判断合法的老代码效率太低了,改了一下之后终于成功的过了。

    这个题其实虽然有些剪枝但是还是好暴力……

    看一下代码。

    // luogu-judger-enable-o2
    #include<cstdio>
    #include<algorithm>
    #include<cstring>
    #include<iostream>
    #include<cmath>
    #include<ctime>
    #define rep(i,a,n) for(int i = a;i <= n;i++)
    #define per(i,n,a) for(int i = n;i >= a;i--)
    #define enter putchar('
    ')
    
    using namespace std;
    typedef long long ll;
    const int M = 1005;
    
    int read()
    {
       int ans = 0,op = 1;
       char ch = getchar();
       while(ch < '0' || ch > '9')
       {
          if(ch == '-') op = -1;
          ch = getchar();
       }
       while(ch >= '0' && ch <= '9')
       {
          ans *= 10;
          ans += ch - '0';
          ch = getchar();
       }
       return ans * op;
    }
    
    bool pd[15][15],flag;
    int num[15][15],visg[11][11],vish[11][11],visl[11][11];
    int curx[15] = {0,1,1,1,4,4,4,7,7,7};
    int cury[15] = {0,1,1,1,4,4,4,7,7,7};
    int sco[11][11] =
    {
       {0,0,0,0,0,0,0,0,0,0},
       {0,6,6,6,6,6,6,6,6,6},
       {0,6,7,7,7,7,7,7,7,6},
       {0,6,7,8,8,8,8,8,7,6},
       {0,6,7,8,9,9,9,8,7,6},
       {0,6,7,8,9,10,9,8,7,6},
       {0,6,7,8,9,9,9,8,7,6},
       {0,6,7,8,8,8,8,8,7,6},
       {0,6,7,7,7,7,7,7,7,6},
       {0,6,6,6,6,6,6,6,6,6}
    };
    
    struct line
    {
       int sum,id,loc;
       bool operator < (const line &g) const
       {
          return sum < g.sum;
       }
    }a[11];
    
    int m,n,tot,px,py,maxn = -1;
    
    bool find(int kx)
    {
       rep(i,a[kx].loc,9)
       {
          rep(j,1,9)
          {
         if(!num[a[i].id][j])
         {
            px = a[i].id,py = j;
            return 1;
         }
          }
       }
       return 0;
    }
    
    void write()
    { 
       rep(i,1,9)
       {
          rep(j,1,9) printf("%d ",num[i][j]);enter;
       }
    }
    
    void calc()
    {
       int cur = 0;
       rep(i,1,9)
          rep(j,1,9) cur += num[i][j] * sco[i][j];
       maxn = max(maxn,cur);
    }
    
    void dfs(int dx,int dy)
    {
       int g = (dx - 1) / 3 * 3 + (dy - 1) / 3;
       rep(k,1,9)
       {
          if(vish[dx][k] || visl[dy][k] || visg[g][k]) continue;
          vish[dx][k] = visl[dy][k] = visg[g][k] = 1,num[dx][dy] = k;
          if(find(dx)) dfs(px,py);
          else calc();//write();
          vish[dx][k] = visl[dy][k] = visg[g][k] = 0,num[dx][dy] = 0;
       }
    }
    int main()
    {
       clock_t start,end;
       start = clock();
       rep(i,1,9)
       {
          a[i].id = i;
          rep(j,1,9)
          {
         int g = (i - 1) / 3 * 3 + (j - 1) / 3;
         num[i][j] = read();
         if(!num[i][j]) a[i].sum++;
         else vish[i][num[i][j]] = visl[j][num[i][j]] = 1,visg[g][num[i][j]] = 1;
          }
       }
       sort(a+1,a+10);
       rep(i,1,9) a[a[i].id].loc = i;
       if(find(a[1].id)) dfs(px,py);
       printf("%d
    ",maxn);
       end = clock();
       double seconds = (double)(end - start) / CLOCKS_PER_SEC;
       //printf("%.8lf
    ",seconds);
       return 0;
    }
         
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  • 原文地址:https://www.cnblogs.com/captain1/p/9880346.html
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