【BZOJ1237】配对（贪心，DP）

题意：有n个a[i]和b[i]，调整顺序使abs(a[i]-b[i])之和最小，但a[i]<>b[i]。保证所有 Ai各不相同，Bi也各不相同。

30%的数据满足：n <= 104 100%的数据满足：1 <= n <= 105，Ai和Bi均为1到106之间的整数。

思路：可以证明调整范围一定在三个数以内

var dp:array[0..110000]of int64;
    a,b,c:array[0..110000]of longint;
    n,i:longint;

function clac(x,y:longint):int64;
begin
 if a[x]=b[y] then exit(1<<60)
  else exit(abs(a[x]-b[y]));
end;

procedure swap(var x,y:longint);
var t:longint;
begin
 t:=x; x:=y; y:=t;
end;

function min(x,y:int64):int64;
begin
 if x<y then exit(x);
 exit(y);
end;

procedure qsort(l,r:longint);
var i,j,mid:longint;
begin
 i:=l; j:=r; mid:=c[(l+r)>>1];
 repeat
  while mid>c[i] do inc(i);
  while mid<c[j] do dec(j);
  if i<=j then
  begin
   swap(c[i],c[j]);
   inc(i); dec(j);
  end;
 until i>j;
 if l<j then qsort(l,j);
 if i<r then qsort(i,r);
end;

begin
 assign(input,'bzoj1237.in'); reset(input);
 assign(output,'bzoj1237.out'); rewrite(output);
 readln(n);
 for i:=1 to n do read(a[i],b[i]);
 for i:=1 to n do c[i]:=a[i];
 qsort(1,n);
 for i:=1 to n do a[i]:=c[i];
 for i:=1 to n do c[i]:=b[i];
 qsort(1,n);
 for i:=1 to n do b[i]:=c[i];
 dp[1]:=clac(1,1);
 dp[2]:=min(clac(1,2)+clac(2,1),dp[1]+clac(2,2));
 for i:=3 to n do
 begin
  dp[i]:=1<<60;
  dp[i]:=min(dp[i],dp[i-1]+clac(i,i));
  dp[i]:=min(dp[i],dp[i-2]+clac(i,i-1)+clac(i-1,i));
  dp[i]:=min(dp[i],dp[i-3]+clac(i-2,i)+clac(i-1,i-2)+clac(i,i-1));
  dp[i]:=min(dp[i],dp[i-3]+clac(i-2,i-1)+clac(i-1,i)+clac(i,i-2));
  dp[i]:=min(dp[i],dp[i-3]+clac(i-2,i)+clac(i-1,i-1)+clac(i,i-2));
 end;
 if dp[n]>(1<<59) then writeln(-1)
  else writeln(dp[n]);
 close(input);
 close(output);
end.