[NOI 1999]棋盘分割(动态规划)

题目就自己看就行了。NOI 1999 day2第一题。没想到day2的题目也可以这么水。

这个题就是一个动态规划，运用记忆化搜索。

dp[x1,y1,x2,y2,k]表示将x1,y1,x2,y2为左上和右下角的矩形分成K份，其中方差没除N的部分的最小值是多少。

方程应该很好想。

【代码】

program chess; var sum,a:array[0..8,0..8] of longint; f:array[1..8,1..8,1..8,1..8] of longint; dp:array[1..8,1..8,1..8,1..8,1..16] of real; i,j,n,p,q:longint; aver:real; function min(x,y:real):real; begin if x<y then exit(x) else exit(y); end; function find(x1,y1,x2,y2,k:integer):real; var ans:real; i:longint; begin if (x1=x2) and (y1=y2) then exit(maxlongint); if dp[x1,y1,x2,y2,k]>0 then exit(dp[x1,y1,x2,y2,k]); if k=1 then exit(sqr(f[x1,y1,x2,y2]-aver)); ans:=maxlongint; for i:=x1 to x2-1 do ans:=min(ans,find(x1,y1,i,y2,k-1)+sqr(f[i+1,y1,x2,y2]-aver)); for i:=x1+1 to x2 do ans:=min(ans,find(i,y1,x2,y2,k-1)+sqr(f[x1,y1,i-1,y2]-aver)); for i:=y1 to y2-1 do ans:=min(ans,find(x1,y1,x2,i,k-1)+sqr(f[x1,i+1,x2,y2]-aver)); for i:=y1+1 to y2 do ans:=min(ans,find(x1,i,x2,y2,k-1)+sqr(f[x1,y1,x2,i-1]-aver)); dp[x1,y1,x2,y2,k]:=ans; exit(ans); end; begin assign(input,'chess.in'); assign(output,'chess.out'); reset(input); rewrite(output); readln(n); for i:=1 to 8 do for j:=1 to 8 do read(a[i,j]); for i:=1 to 8 do for j:=1 to 8 do sum[i,j]:=sum[i-1,j]+sum[i,j-1]+a[i,j]-sum[i-1,j-1]; for i:=1 to 8 do for j:=1 to 8 do for p:=i to 8 do for q:=j to 8 do f[i,j,p,q]:=sum[p,q]-sum[i-1,q]-sum[p,j-1]+sum[i-1,j-1]; aver:=sum[8,8]/n; writeln(sqrt(find(1,1,8,8,n)/n):0:3); close(input); close(output); end.