hdu 3333 Turing Tree(主席树)

Turing Tree

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6004    Accepted Submission(s): 2152


Problem Description
After inventing Turing Tree, 3xian always felt boring when solving problems about intervals, because Turing Tree could easily have the solution. As well, wily 3xian made lots of new problems about intervals. So, today, this sick thing happens again...

Now given a sequence of N numbers A1, A2, ..., AN and a number of Queries(i, j) (1≤i≤j≤N). For each Query(i, j), you are to caculate the sum of distinct values in the subsequence Ai, Ai+1, ..., Aj.
 

Input
The first line is an integer T (1 ≤ T ≤ 10), indecating the number of testcases below.
For each case, the input format will be like this:
* Line 1: N (1 ≤ N ≤ 30,000).
* Line 2: N integers A1, A2, ..., AN (0 ≤ Ai ≤ 1,000,000,000).
* Line 3: Q (1 ≤ Q ≤ 100,000), the number of Queries.
* Next Q lines: each line contains 2 integers i, j representing a Query (1 ≤ i ≤ j ≤ N).
 

Output
For each Query, print the sum of distinct values of the specified subsequence in one line.
 

Sample Input

2 3 1 1 4 2 1 2 2 3 5 1 1 2 1 3 3 1 5 2 4 3 5
 

Sample Output

1 5 6 3 6
 

Author
3xian@GDUT
 

题意:

给了N个数和M次查询,每次查询L到R区间内不同数的和。

思路:

用主席树更新N个版本,树上每个节点保留的是区间里面不同数的和。

网上主席树的板子都不一样,发现了这个很好用的板子,写起来顺手,就当作模板了。

代码:

#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;
const int maxn=30005;
int cnt,root[maxn],a[maxn],m;
mapmp;
struct node
{
    int l,r,ls,rs;
    long long sum;
}tr[maxn*40];
vectorv;
int getid(int x)
{
    return lower_bound(v.begin(),v.end(),x)-v.begin()+1;
}
int build(int l,int r)
{
    int num=++cnt;
    tr[num].l=l,tr[num].r=r,tr[num].sum=0;
    if(l==r) return num;
    int mid=(l+r)/2;
    tr[num].ls=build(l,mid);
    tr[num].rs=build(mid+1,r);
    return num;
}
int update(int pre,int pos,int v)
{
    int num=++cnt;
    tr[num]=tr[pre];
    tr[num].sum+=(long long)v;
    int l=tr[num].l,r=tr[num].r;
    if(l==r) return num;
    int mid=(l+r)/2;
    if(pos<=mid) tr[num].ls=update(tr[pre].ls,pos,v);
    else tr[num].rs=tr[num].rs=update(tr[pre].rs,pos,v);
    return num;
}
long long query(int num,int L,int R)
{
    int l=tr[num].l,r=tr[num].r;
    if(L<=l&&r<=R)
    {
        return tr[num].sum;
    }
    long long sum=0;
    int mid=(l+r)/2;
    if(L<=mid) sum+=query(tr[num].ls,L,R);
    if(R>mid) sum+=query(tr[num].rs,L,R);
    return sum;
}
void init()
{
    cnt=0;
    v.clear();
    mp.clear();
}
int main()
{
    int t,n,l,r;
    scanf("%d",&t);
    while(t--)
    {
        init();
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            v.push_back(a[i]);
        }
        sort(v.begin(),v.end());
        v.erase(unique(v.begin(),v.end()),v.end());
        root[0]=build(1,n);
        for(int i=1;i<=n;i++)
        {
            int id=getid(a[i]);
            if(mp[id]==0)
            {
                root[i]=update(root[i-1],i,a[i]);
            }
            else
            {
                int t=update(root[i-1],mp[id],-a[i]);
                root[i]=update(t,i,a[i]);
            }
            mp[id]=i;
        }
        scanf("%d",&m);
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d",&l,&r);
            printf("%lld\n",query(root[r],l,r));
        }
    }
    return 0;
}


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