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這道題比我想象之中要暴力一些。

自己一開始有一份寫9*9數獨的代碼，自己試了一下直接交上去只有40分。看來這樣是肯定不行的。考慮優化，想在中間貪貪心啥的，但是難以保證正確性。最後學了一招，從數字比較多的行開始搜索，這樣會使搜索樹變得小一些，時間就會減少一些。

然後交上去之後還是只有75分……發現自己數獨判斷合法的老代碼效率太低了，改了一下之後終於成功的過了。

這個題其實雖然有些剪枝但是還是好暴力……

看一下代碼。

// 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(‘\n‘)

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\n",maxn);
   end = clock();
   double seconds = (double)(end - start) / CLOCKS_PER_SEC;
   //printf("%.8lf\n",seconds);
   return 0;
}
     

NOIP2009 靶型數獨