bzoj2298

不难发现必然是两个人之间话产生矛盾或自身话有问题
很显然，当ai>aj时，若ai<n-bj时i,j两人话矛盾
ai<ai<n-bj，这东西有没有数轴上的线段的既视感？
我们只要求出做多不相交的线段就可得到最少谎话数

 var a,b,f:array[..] of longint;
     i,j,k,n:longint;

 function min(a,b:longint):longint;
   begin
     if a>b then exit(b) else exit(a);
   end;

 function max(a,b:longint):longint;
   begin
     if a>b then exit(a) else exit(b);
   end;

 procedure swap(var a,b:longint);
   var c:longint;
   begin
     c:=a;
     a:=b;
     b:=c;
   end;

 procedure sort(l,r:longint);
   var i,j,x,y:longint;
   begin
     i:=l;
     j:=r;
     x:=b[(l+r) shr ];
     y:=a[(l+r) shr ];
     repeat
       while (b[i]<x) or (b[i]=x) and (a[i]<y) do inc(i);
       while (x<b[j]) or (b[j]=x) and (y<a[j]) do dec(j);
       if not(i>j) then
       begin
         swap(a[i],a[j]);
         swap(b[i],b[j]);
         inc(i);
         dec(j);
       end;
     until i>j;
     if l<j then sort(l,j);
     if i<r then sort(i,r);
   end;

 begin
   readln(n);
   for i:= to n do
   begin
     readln(a[i],b[i]);
     b[i]:=n-b[i];
   end;
   sort(,n);
   j:=;
   for i:= to n do
   begin
     f[i]:=f[i-];
     while b[j]=i do
     begin
       k:=j+;
       while (a[k]=a[j]) and (b[k]=b[j]) do inc(k);
       f[i]:=max(f[i],f[a[j]]+min(k-j,b[j]-a[j]));  //注意这个地方相同分数的数目
       j:=k;
     end;
   end;
   writeln(n-f[n]);
 end.