hdu 3333 Turing Tree

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Turing Tree

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 3747    Accepted Submission(s): 1271

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

先按查询的右坐标升序排列。把每个数放进线段树的时候，先判断再之前他有没有在线段树，如果在，则删除它，并把他的位置更新到当前点

#include<iostream>#include<stdio.h>#include<string>#include<math.h>#include<vector>#include<queue>#include<map>#include<string.h>#include<algorithm>#define N 300005#define mod 10007#define ll __int64#define pii pair<int,int>//#pragma comment(linker, "/STACK:1024000000,1024000000")#define ex 2.7182818284590452354+#define pi 3.141592653589793239#define eps 1e-8#define INFF 999999999using namespace std;ll a[N],ans[N];map<ll,int>vis;struct T{    int l,r,id;}op[N<<2];bool cmp(T a,T b){    return a.r<b.r;}struct node{    int l,r;    ll sum;    int mid()    {        return (l+r)>>1;    }}t[N<<2];void up(node &fa,node &lson,node &rson){    fa.sum=lson.sum+rson.sum;}void push_up(int root){    up(t[root],t[root<<1],t[root<<1|1]);}void build(int root,int l,int r){    t[root].l=l;    t[root].r=r;    t[root].sum=0;    if(l==r)        return;    int mid=t[root].mid();    build(root<<1,l,mid);    build(root<<1|1,mid+1,r);}void update(int root,int pos,ll val){    if(t[root].l==t[root].r)    {        t[root].sum+=val;        return;    }    int mid=t[root].mid();    if(pos<=mid)        update(root<<1,pos,val);    else        update(root<<1|1,pos,val);    push_up(root);}node query(int root,int l,int r){    if(l<=t[root].l&&t[root].r<=r)        return t[root];    int mid=t[root].mid();    node temp[3];    int xx=0;    if(l<=mid)    {        temp[1]=query(root<<1,l,r);        xx++;    }    if(r>=mid+1)    {        temp[2]=query(root<<1|1,l,r);        xx+=2;    }    if(xx<3)        return temp[xx];    up(temp[0],temp[1],temp[2]);    return temp[0];}int main(){    int tt;    while(scanf("%d",&tt)!=EOF)    {        while(tt--)        {            vis.clear();            int i,k,n,m;            scanf("%d",&n);            for(i=1;i<=n;i++)                scanf("%I64d",&a[i]);            build(1,1,n);            scanf("%d",&m);            for(i=0;i<m;i++)            {                scanf("%d %d",&op[i].l,&op[i].r);                op[i].id=i;            }            sort(op,op+m,cmp);            k=1;            for(i=0;i<m;i++)            {                for(;k<=op[i].r;k++)                {                    if(vis[a[k]]!=0)                        update(1,vis[a[k]],-a[k]);                    vis[a[k]]=k;                    update(1,k,a[k]);                }                ans[op[i].id]=query(1,op[i].l,op[i].r).sum;            }            for(i=0;i<m;i++)                printf("%I64d\n",ans[i]);        }    }    return 0;}