HDU 3333 Turing Tree

HDU 3333 Turing Tree

和HDU 3874类似，只不过数字的范围大了不少，不能直接开数组记录每个数字最后出现的位置，使用map<int,int>记录。
以下是我的代码：

#include < iostream >
#include < map >
#include < algorithm >
#include < cstdio >
#include < cstring >
#define  lowbit(x) ((x)&(-(x)))
using   namespace  std;
const   int  kMaxn( 30007 );
const   int  kMaxm( 100007 );

struct  Interval
{
     int  a,b;
     int  id;
};
bool   operator < ( const  Interval  & a, const  Interval  & b)
{
     return  (a.b < b.b);
}

int  n,m,r[kMaxn];
Interval query[kMaxm];
long   long  bit[kMaxn],ans[kMaxm];

void  Add( int  x, int  delta)
{
     for ( int  i = x;i <= n;i += lowbit(i))
        bit[i] += delta;
}

long   long  Sum( int  x)
{
     long   long  re( 0 );
     for ( int  i = x;i > 0 ;i -= lowbit(i))
        re += bit[i];
     return  re;
}

int  main()
{
     // freopen("data.in","r",stdin);

     int  T;
    scanf( " %d " , & T);
     while (T -- )
    {
        scanf( " %d " , & n);
         for ( int  i = 1 ;i <= n;i ++ )
            scanf( " %d " , & r[i]);
        scanf( " %d " , & m);
         for ( int  i = 1 ;i <= m;i ++ )
        {
            scanf( " %d%d " , & query[i].a, & query[i].b);
            query[i].id = i;
        }
         //   Input
        
        sort(query + 1 ,query + m + 1 );
        memset(bit, 0 , sizeof (bit));
        
        map < int , int >  pos;
         for ( int  i = 1 ,j = 1 ;i <= n  &&  j <= m;i ++ )
        {
            Add(i,r[i]);
             if (pos.count(r[i]))
                Add(pos[r[i]], - r[i]);
            pos[r[i]] = i;

             while (i == query[j].b)
            {
                ans[query[j].id] = Sum(query[j].b) - Sum(query[j].a - 1 );
                j ++ ;
            }
        }

         for ( int  i = 1 ;i <= m;i ++ )
            printf( " %I64d\n " ,ans[i]);
    }

     return   0 ;
}