Description
将一个a*b的数字矩阵进行如下分割:将原矩阵沿某一条直线分割成两个矩阵,再将生成的两个矩阵继续如此分割(当然也可以只分割其中的一个),这样分割了(n-1)次后,原矩阵被分割成了n个矩阵。(每次分割都只能沿着数字间的缝隙进行)原矩阵中每一位置上有一个分值,一个矩阵的总分为其所含各位置上分值之和。现在需要把矩阵按上述规则分割成n个矩阵,并使各矩阵总分的均方差最小。请编程对给出的矩阵及n,求出均方差的最小值。
Input
第一行为3个整数,表示a,b,n(1
Output
仅一个数,为均方差的最小值(四舍五入精确到小数点后2位)
Sample Input
5 4 4
2 3 4 6
5 7 5 1
10 4 0 5
2 0 2 3
4 1 1 1
Sample Output
0.50
记忆化搜索,f[i,j,k,l,n]表示矩形(i,j,k,l)分成n块最小方差(最后再除以n,一开始没除,我真的是个傻×,方差都不会算)
1 const 2 inf=999999999999; 3 var 4 f:array[0..10,0..10,0..10,0..10,0..10]of double; 5 aa:array[0..10,0..10]of double; 6 a,b,n:longint; 7 sum:double; 8 9 function min(x,y:double):double; 10 begin 11 if x<y then exit(x); 12 exit(y); 13 end; 14 15 function fn(x1,x2,y1,y2,k:longint):double; 16 var 17 i,j:longint; 18 begin 19 if f[x1,x2,y1,y2,k]<inf then exit(f[x1,x2,y1,y2,k]); 20 if (x2-x1+1)*(y2-y1+1)<k then exit(inf); 21 for i:=x1 to x2-1 do 22 for j:=1 to k-1 do 23 f[x1,x2,y1,y2,k]:=min(f[x1,x2,y1,y2,k],fn(x1,i,y1,y2,j)+fn(i+1,x2,y1,y2,k-j)); 24 for i:=y1 to y2-1 do 25 for j:=1 to k-1 do 26 f[x1,x2,y1,y2,k]:=min(f[x1,x2,y1,y2,k],fn(x1,x2,y1,i,j)+fn(x1,x2,i+1,y2,k-j)); 27 exit(f[x1,x2,y1,y2,k]); 28 end; 29 30 procedure main; 31 var 32 i,j,k,l,r:longint; 33 begin 34 read(a,b,n); 35 for i:=1 to a do 36 for j:=1 to b do 37 begin 38 read(aa[i,j]); 39 sum:=sum+aa[i,j]; 40 aa[i,j]:=aa[i,j]+aa[i-1,j]+aa[i,j-1]-aa[i-1,j-1]; 41 end; 42 sum:=sum/n; 43 for i:=1 to a do 44 for j:=i to a do 45 for k:=1 to b do 46 for l:=k to b do 47 for r:=1 to n do 48 f[i,j,k,l,r]:=inf; 49 for i:=1 to a do 50 for j:=i to a do 51 for k:=1 to b do 52 for l:=k to b do 53 f[i,j,k,l,1]:=sqr(aa[j,l]+aa[i-1,k-1]-aa[j,k-1]-aa[i-1,l]-sum); 54 write(sqrt(fn(1,a,1,b,n)/n):0:2); 55 end; 56 57 begin 58 main; 59 end.