HDU-3333-Turing Tree-（树状数组，离散化）

Turing Tree

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5579    Accepted Submission(s): 1991

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

231 1 421 22 351 1 2 1 331 52 43 5

 

Sample Output

15636

 

Author

3xian@GDUT

 

Source

HDOJ Monthly Contest – 2010.03.06

 

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与题目hdu-3874-Necklace-（树状数组）相同，只是这里数据更大，需要借用STL的函数离散化处理，

lower_bound(begin, end, value); 返回>=value的元素的第一个位置。函数lower_bound()在first和last中的前闭后开区间进行二分查找，返回大于或等于val的第一个元素位置。如果所有元素都小于val，则返回last的位置

介绍lower_bound()的文章：http://blog.csdn.net/niushuai666/article/details/6734403

代码：

#include<iostream>#include<string>#include<cstdio>#include<algorithm>#include<cmath>#include<iomanip>#include<queue>#include<cstring>#include<map>using namespace std;typedef long long ll;int n,m;ll tree[30005],ans[100005];int a[30005],vis[30005],b[30005],pos[30005];struct node{    int s,e,id;    bool operator <(const node& b) const    {        return e<b.e||(e==b.e&&s<b.s);    }}p[100005];int lowbit(int i){    return i&(-i);}void add(int i,int v){    while(i<=n)    {        tree[i]+=v;        i+=lowbit(i);    }}ll sum(int i)  //一定注意i>=1,切记i不能为0{    ll res=0;    while(i>0)    {        res+=tree[i];        i-=lowbit(i);    }    return res;}int main(){    int i,T;    scanf("%d",&T);    while(T--)    {        scanf("%d",&n);        memset(tree,0,sizeof(tree));        memset(vis,0,sizeof(vis));        for(i=1;i<=n;i++)        {            scanf("%d",&a[i]);            b[i]=a[i];        }        scanf("%d",&m);        for(i=1;i<=m;i++)        {            scanf("%d%d",&p[i].s,&p[i].e);            p[i].id=i;        }        sort(p+1,p+m+1);        sort(b+1,b+n+1);        for(i=1;i<=n;i++)        {            pos[i]=lower_bound(b+1,b+n+1,a[i])-b;        }        int it=1;        for(i=1;i<=m;i++)        {            while(it<=p[i].e)            {                if(vis[pos[it]]!=0)                {                    add(vis[pos[it]],-a[it]);                }                vis[pos[it]]=it;                add(it,a[it]);                it++;            }            ans[p[i].id]=sum(p[i].e)-sum(p[i].s-1);        }        for(i=1;i<=m;i++)        {            printf("%I64d\n",ans[i]);        }    }    return 0;}

参考博客：http://blog.csdn.net/just_water/article/details/9716983