【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=100000;
var dp:array[0..1,-6100..6100]of longint;
    a,b:array[1..1100]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:=1 to n do read(a[i],b[i]);
 v:=0;
 fillchar(dp[v],sizeof(dp[v]),$1f);
 dp[0,0]:=0;
 for i:=0 to n-1 do
 begin
  v:=1-v; fillchar(dp[v],sizeof(dp[v]),$1f);
  for j:=-5*i to 5*i do
   if dp[1-v,j]<oo then
   begin
    t:=j+a[i+1]-b[i+1];
    dp[v,t]:=min(dp[v,t],dp[1-v,j]);
    t:=j+b[i+1]-a[i+1];
    dp[v,t]:=min(dp[v,t],dp[1-v,j]+1);
   end;
//  print;
 end;
 ans:=oo;
 for i:=0 to 5*n do
 begin
  ans:=min(ans,min(dp[v,i],dp[v,-i]));
  if ans<oo then break;
 end;
 writeln(ans);
 
end.