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('\n')

using namespace std;
typedef long long ll;
const int M = ;

int read()
{
   int ans = ,op = ;
   char ch = getchar();
   while(ch < '' || ch > '')
   {
      if(ch == '-') op = -;
      ch = getchar();
   }
   while(ch >= '' && ch <= '')
   {
      ans *= ;
      ans += ch - '';
      ch = getchar();
   }
   return ans * op;
}

bool pd[][],flag;
int num[][],visg[][],vish[][],visl[][];
int curx[] = {,,,,,,,,,};
int cury[] = {,,,,,,,,,};
int sco[][] =
{
   {,,,,,,,,,},
   {,,,,,,,,,},
   {,,,,,,,,,},
   {,,,,,,,,,},
   {,,,,,,,,,},
   {,,,,,,,,,},
   {,,,,,,,,,},
   {,,,,,,,,,},
   {,,,,,,,,,},
   {,,,,,,,,,}
};

struct line
{
   int sum,id,loc;
   bool operator < (const line &g) const
   {
      return sum < g.sum;
   }
}a[];

int m,n,tot,px,py,maxn = -;

bool find(int kx)
{
   rep(i,a[kx].loc,)
   {
      rep(j,,)
      {
     if(!num[a[i].id][j])
     {
        px = a[i].id,py = j;
        return ;
     }
      }
   }
   return ;
}

void write()
{
   rep(i,,)
   {
      rep(j,,) printf("%d ",num[i][j]);enter;
   }
}

void calc()
{
   int cur = ;
   rep(i,,)
      rep(j,,) cur += num[i][j] * sco[i][j];
   maxn = max(maxn,cur);
}

void dfs(int dx,int dy)
{
   int g = (dx - ) /  *  + (dy - ) / ;
   rep(k,,)
   {
      if(vish[dx][k] || visl[dy][k] || visg[g][k]) continue;
      vish[dx][k] = visl[dy][k] = visg[g][k] = ,num[dx][dy] = k;
      if(find(dx)) dfs(px,py);
      else calc();//write();
      vish[dx][k] = visl[dy][k] = visg[g][k] = ,num[dx][dy] = ;
   }
}
int main()
{
   clock_t start,end;
   start = clock();
   rep(i,,)
   {
      a[i].id = i;
      rep(j,,)
      {
     int g = (i - ) /  *  + (j - ) / ;
     num[i][j] = read();
     if(!num[i][j]) a[i].sum++;
     else vish[i][num[i][j]] = visl[j][num[i][j]] = ,visg[g][num[i][j]] = ;
      }
   }
   sort(a+,a+);
   rep(i,,) a[a[i].id].loc = i;
   if(find(a[].id)) dfs(px,py);
   printf("%d\n",maxn);
   end = clock();
   double seconds = (double)(end - start) / CLOCKS_PER_SEC;
   //printf("%.8lf\n",seconds);
   return ;
}