【Vijos1222】等值拉面（DP）

题意：有N个数对(a[i],b[i])

每次可以把(x,y)变成(x+a[i],y+b[i])或(x+b[i],x+a[i])，后者称为交换一次

求使abs(x-y)最小时的最小交换次数

n<=1000 1<=a[i],b[i]<=6

思路：设dp[i,j]为前i个数对，使x-y=j时的最小交换次数 转移即可

 const oo=;
 var dp:array[..,-..]of longint;
     a,b:array[..]of longint;
     n,i,v,j,t,ans:longint;
 
 function min(x,y:longint):longint;
 begin
  if x<y then exit(x);
  exit(y);
 end;
 
 begin
 
  readln(n);
  for i:= to n do read(a[i],b[i]);
  v:=;
  fillchar(dp[v],sizeof(dp[v]),$1f);
  dp[,]:=;
  for i:= to n- do
  begin
   v:=-v; fillchar(dp[v],sizeof(dp[v]),$1f);
   for j:=-*i to *i do
    if dp[-v,j]<oo then
    begin
     t:=j+a[i+]-b[i+];
     dp[v,t]:=min(dp[v,t],dp[-v,j]);
     t:=j+b[i+]-a[i+];
     dp[v,t]:=min(dp[v,t],dp[-v,j]+);
    end;
 //  print;
  end;
  ans:=oo;
  for i:= to *n do
  begin
   ans:=min(ans,min(dp[v,i],dp[v,-i]));
   if ans<oo then break;
  end;
  writeln(ans);
 
 end.