BZOJ1048: [HAOI2007]分割矩阵

题目：http://www.lydsy.com/JudgeOnline/problem.php?id=1048

题解：搞清题意之后来个记忆化爆搜就行了。

代码：

 #include<cstdio>

 #include<cstdlib>

 #include<cmath>

 #include<cstring>

 #include<algorithm>

 #include<iostream>

 #include<vector>

 #include<map>

 #include<set>

 #include<queue>

 #include<string>

 #define inf 1000000000

 #define maxn 11

 #define maxm 200000+5

 #define eps 1e-10

 #define ll long long

 #define pa pair<int,int>

 #define for0(i,n) for(int i=0;i<=(n);i++)

 #define for1(i,n) for(int i=1;i<=(n);i++)

 #define for2(i,x,y) for(int i=(x);i<=(y);i++)

 #define for3(i,x,y) for(int i=(x);i>=(y);i--)

 #define for4(i,x) for(int i=head[x],y=e[i].go;i;i=e[i].next,y=e[i].go)

 #define for5(n,m) for(int i=1;i<=n;i++)for(int j=1;j<=m;j++)

 #define mod 1000000007
 #define sqr(x) (x)*(x)

 using namespace std;

 inline int read()

 {

     int x=,f=;char ch=getchar();

     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}

     while(ch>=''&&ch<=''){x=*x+ch-'';ch=getchar();}

     return x*f;

 }
 int n,m,k,s[maxn][maxn];
 double ave,f[maxn][maxn][maxn][maxn][maxn];
 inline double dp(int x1,int y1,int x2,int y2,int z)
 {
     double &t=f[x1][y1][x2][y2][z];
     if(t<inf)return t;
     if(!z)return t=sqr(s[x2][y2]-s[x1-][y2]-s[x2][y1-]+s[x1-][y1-]-ave);
     t=inf;
     for0(i,z-)
      {
          for2(j,x1,x2-)t=min(t,dp(x1,y1,j,y2,i)+dp(j+,y1,x2,y2,z--i));
          for2(j,y1,y2-)t=min(t,dp(x1,y1,x2,j,i)+dp(x1,j+,x2,y2,z--i));
      }
     //cout<<x1<<' '<<y1<<' '<<x2<<' '<<y2<<' '<<z<<' '<<t<<endl;
     return t;
 }

 int main()

 {

     freopen("input.txt","r",stdin);

     freopen("output.txt","w",stdout);

     n=read();m=read();k=read();
     for1(i,n)for1(j,m)s[i][j]=s[i-][j]+s[i][j-]-s[i-][j-]+read();
     ave=(double)s[n][m]/(double)k;
     for1(i1,n)for1(i2,m)for1(i3,n)for1(i4,m)for0(i5,k)f[i1][i2][i3][i4][i5]=inf;
     dp(,,n,m,k-);
     printf("%.2f\n",sqrt((double)f[][][n][m][k-]/(double)k));

     return ;

 }