题目

Source

http://acm.hdu.edu.cn/showproblem.php?pid=3333

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

分析

题目大概说给一个序列，多次询问一个区间内不同数之和。

图灵树的来历原来是这个。。经典的线段树离线所有查询右端点排序的题目吧。。

所有询问右端点排序后，从小到大扫过去，线段树维护序列区间和，用一个map记录各个数最右边出现的位置，一遇到一个数就把之前位置消除并更新当前位置，相当于把各个数尽量向右移动。。

代码

#include<cstdio>#include<cstring>#include<map>#include<algorithm>using namespace std;#define MAXN 33333struct Query{	int i,l,r;	bool operator<(const Query &q) const{		return r<q.r;	}}que[111111];long long tree[MAXN<<2];int N,x,y;void update(int i,int j,int k){	if(i==j){		tree[k]+=y;		return;	}	int mid=i+j>>1;	if(x<=mid) update(i,mid,k<<1);	else update(mid+1,j,k<<1|1);	tree[k]=tree[k<<1]+tree[k<<1|1];}long long query(int i,int j,int k){	if(x<=i && j<=y) return tree[k];	int mid=i+j>>1;	long long ret=0;	if(x<=mid) ret+=query(i,mid,k<<1);	if(y>mid) ret+=query(mid+1,j,k<<1|1);	return ret;}int a[MAXN];long long ans[111111];int main(){	int t,n,m;	scanf("%d",&t);	while(t--){		scanf("%d",&n);		for(int i=1; i<=n; ++i){			scanf("%d",a+i);		}		scanf("%d",&m);		for(int i=0; i<m; ++i){			scanf("%d%d",&que[i].l,&que[i].r);			que[i].i=i;		}		sort(que,que+m);		map<int,int> posi;		memset(tree,0,sizeof(tree));		for(N=1; N<n; N<<=1);		int p=0;		for(int i=0; i<m; ++i){			while(p<que[i].r){				++p;				if(posi.count(a[p])){					x=posi[a[p]]; y=-a[p];					update(1,N,1);				}				x=p; y=a[p];				update(1,N,1);				posi[a[p]]=p;			}			x=que[i].l; y=que[i].r;			ans[que[i].i]=query(1,N,1);		}		for(int i=0; i<m; ++i){			printf("%lld\n",ans[i]);		}	}	return 0;}

HDU3333 Turing Tree（线段树）