PKU1190 post office

       好久不做题了，今天特用这道DP释怀一下。

       应该不算洪水般的裸DP了，需要想几分钟…

       首先初始化出在任意两点间建立一个邮局时这两点间村庄所用距离的最小值。接着开二维数组表示前j 个村庄的人去前 i 个邮局所用的最小距离和。

       这样就应该搞定了。感兴趣的呢，可以滚动一下，或者四边形一下！

program post_office;const inf=200000000;var dist:array[0..301,0..301] of longint; f:array[0..31,0..301] of longint; data:array[0..301] of longint; i,j,k,key,n,m:longint;function getmin(a,b:longint):longint;begin if a<b then exit(a); exit(b);end;begin assign(input,'a.in'); reset(input); assign(output,'a.out'); rewrite(output); readln(n,m); for i:=1 to n do read(data[i]); fillchar(dist,sizeof(dist),0); for i:=1 to n do for j:=i to n do begin key:=data[(i+j) shr 1]; for k:=i to j do inc(dist[i,j],abs(data[k]-key)); end; filldword(f,sizeof(f) shr 2,inf); for j:=1 to n do f[1,j]:=dist[1,j]; for i:=2 to m do for j:=i to n do for k:=i-1 to j-1 do f[i,j]:=getmin(f[i,j],dist[k+1,j]+f[i-1,k]); writeln(f[m,n]); close(input); close(output);end.