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：
1 #include<cstdio> 2 3 #include<cstdlib> 4 5 #include<cmath> 6 7 #include<cstring> 8 9 #include<algorithm> 10 11 #include<iostream> 12 13 #include<vector> 14 15 #include<map> 16 17 #include<set> 18 19 #include<queue> 20 21 #include<string> 22 23 #define inf 1000000000 24 25 #define maxn 11 26 27 #define maxm 200000+5 28 29 #define eps 1e-10 30 31 #define ll long long 32 33 #define pa pair<int,int> 34 35 #define for0(i,n) for(int i=0;i<=(n);i++) 36 37 #define for1(i,n) for(int i=1;i<=(n);i++) 38 39 #define for2(i,x,y) for(int i=(x);i<=(y);i++) 40 41 #define for3(i,x,y) for(int i=(x);i>=(y);i--) 42 43 #define for4(i,x) for(int i=head[x],y=e[i].go;i;i=e[i].next,y=e[i].go) 44 45 #define for5(n,m) for(int i=1;i<=n;i++)for(int j=1;j<=m;j++) 46 47 #define mod 1000000007 48 #define sqr(x) (x)*(x) 49 50 using namespace std; 51 52 inline int read() 53 54 { 55 56 int x=0,f=1;char ch=getchar(); 57 58 while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} 59 60 while(ch>='0'&&ch<='9'){x=10*x+ch-'0';ch=getchar();} 61 62 return x*f; 63 64 } 65 int n,m,k,s[maxn][maxn]; 66 double ave,f[maxn][maxn][maxn][maxn][maxn]; 67 inline double dp(int x1,int y1,int x2,int y2,int z) 68 { 69 double &t=f[x1][y1][x2][y2][z]; 70 if(t<inf)return t; 71 if(!z)return t=sqr(s[x2][y2]-s[x1-1][y2]-s[x2][y1-1]+s[x1-1][y1-1]-ave); 72 t=inf; 73 for0(i,z-1) 74 { 75 for2(j,x1,x2-1)t=min(t,dp(x1,y1,j,y2,i)+dp(j+1,y1,x2,y2,z-1-i)); 76 for2(j,y1,y2-1)t=min(t,dp(x1,y1,x2,j,i)+dp(x1,j+1,x2,y2,z-1-i)); 77 } 78 //cout<<x1<<' '<<y1<<' '<<x2<<' '<<y2<<' '<<z<<' '<<t<<endl; 79 return t; 80 } 81 82 int main() 83 84 { 85 86 freopen("input.txt","r",stdin); 87 88 freopen("output.txt","w",stdout); 89 90 n=read();m=read();k=read(); 91 for1(i,n)for1(j,m)s[i][j]=s[i-1][j]+s[i][j-1]-s[i-1][j-1]+read(); 92 ave=(double)s[n][m]/(double)k; 93 for1(i1,n)for1(i2,m)for1(i3,n)for1(i4,m)for0(i5,k)f[i1][i2][i3][i4][i5]=inf; 94 dp(1,1,n,m,k-1); 95 printf("%.2f\n",sqrt((double)f[1][1][n][m][k-1]/(double)k)); 96 97 return 0; 98 99 }
