SPOJ DQUERY树状数组离线or主席树

D-query

Time Limit: 227MS

Memory Limit: 1572864KB

64bit IO Format: %lld & %llu

Submit Status

Description

English

Vietnamese

Given a sequence of n numbers a1, a2, ..., an and a number of d-queries. A d-query is a pair (i, j) (1 ≤ i ≤ j ≤ n). For each d-query (i, j), you have to return the number of distinct elements in the subsequence ai, ai+1, ..., aj.

Input

Line 1: n (1 ≤ n ≤ 30000).

Line 2: n numbers a1, a2, ..., an (1 ≤ ai ≤ 106).

Line 3: q (1 ≤ q ≤ 200000), the number of d-queries.

In the next q lines, each line contains 2 numbers i, j representing a d-query (1 ≤ i ≤ j ≤ n).

Output

For each d-query (i, j), print the number of distinct elements in the subsequence ai, ai+1, ..., aj in a single line.

Example

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

Output
3
2
3

树状数组的方法：

/*
SPOJ DQUERY(hdu3333)线段树or树状数组离线
查询区间内不同数的个数
本来是学习主席树的，发现这方法不会也就写了下，感觉很机智
先将所有查询按区间右端从小到大排序，如果一个数已经出现过就先把以前位置上的删
掉然后在新的位置上插入，像这样在[l,r]中重复的就只计算了一次
hhh-2016-02-18 14:47:11
*/
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <queue>
#include <vector>
using namespace std;
typedef long long ll;
typedef long double ld;
int tot;
map<int , int >mp;
const int maxn = 100010;
int n;
int a[maxn],ans[maxn*2];
int s[maxn];

struct node
{
int l,r,id;
} qu[maxn*2];

bool cmp(node a,node b)
{
return a.r < b.r;
}

int lowbit(int x)
{
return x&(-x);
}

void add(int x,int val)
{
for(int i = x ; i <= n; i+=lowbit(i))
s[i] += val;
}

int query(int x)
{
int sum = 0;
for(int i = x; i > 0; i -= lowbit(i))
sum += s[i];
return sum;
}

int main()
{
while(scanf("%d",&n) != EOF)
{
mp.clear();
memset(s,0,sizeof(s));
for(int i =1 ; i <= n; i++)
scanf("%d",&a[i]);

int q;
scanf("%d",&q);
for(int i = 1; i <= q; i++)
{
scanf("%d%d",&qu[i].l,&qu[i].r);
qu[i].id = i;
}
sort(qu+1,qu+q+1,cmp);
int t = 1;
for(int i= 1;i <= q;i++)
{
for(;t <= qu[i].r;t++)
{
if(mp[a[t]] != 0) add(mp[a[t]],-1);
mp[a[t]] = t;
add(t,1);
}
ans[qu[i].id] = query(qu[i].r) - query(qu[i].l-1);
}
for(int i = 1;i <= q;i++)
{
printf("%d\n",ans[i]);
}
}
return 0;
}

主席树：  //参考：关于主席树的读书笔记

将1-i用线段树处理，每个表示前i个数的情况。如果每棵线段都建完整的话肯定会mle，我们发现对于前缀[1,i]和前缀[1,i+1]的线段树，如果b[i+1]<=mid (b[i+1]表示a[i+1]离散后的标记) 那么线段树i和线段树i+1的左边是完全相同的，根本不需要在建，只需要用指针指一下就好

//忘了在(=@__@=)哪里看的了

换一种解释，如果我们要修改左子树，那么右子树上的与上一个线段树相比不会变化，只需要指一下就好

/*
主席树 SPOJ DQUERY
查询区间有多少个不同的数。类似于之前树状数组离线的思路，在插入之前先进行判断
如果已经有了，把以前的先删掉再进行插入
hhh-2016-02-18 15:37:48
*/

#include <functional>
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <map>
#include <cmath>
using namespace std;
typedef long long ll;
typedef long double ld;

using namespace std;
const int maxn = 100010;
int tot;
int n;
int a[maxn],t[maxn];
int T[maxn],lson[maxn*30],rson[maxn*30],c[maxn*30];

int build(int l,int r)
{
int root = tot++;
c[root] = 0;
if(l != r)
{
int mid = (l+r)>>1;
lson[root] = build(l,mid);
rson[root] = build(mid+1,r);
}
return root;
}

//如果那里发生改变则兴建一个节点而非像平常修改那个节点的值
int update(int root,int pos,int val)
{
int newroot = tot++;
int tmp = newroot;
c[newroot] = c[root] + val;
int l = 1,r = n;
while(l < r)
{
int mid = (l+r)>>1;
if(pos <= mid)
{
lson[newroot] = tot++;
rson[newroot] = rson[root];
newroot = lson[newroot];
root = lson[root];
r = mid;
}
else
{
lson[newroot] = lson[root];
rson[newroot] = tot++;
newroot = rson[newroot];
root = rson[root];
l = mid+1;
}
c[newroot] = c[root] + val;
}
return tmp;
}

int query(int root,int pos)
{
int l = 1, r = n;
int ans = 0;
while(pos > l)
{
int mid = (l+r)>>1;
if(pos <= mid)
{
ans += c[rson[root]];
r = mid;
root = lson[root];
}
else
{
l = mid+1;
root = rson[root];
}
}
return ans+c[root];
}

int main()
{
while(scanf("%d",&n) !=EOF)
{
tot = 0;
map<int,int> mp;
for(int i = 1; i <= n; i++)
scanf("%d",&a[i]);
T[0] = build(1,n);
for(int i =1; i <= n; i++)
{
if(mp.find(a[i]) == mp.end())
T[i] = update(T[i-1],i,1);
else
{
int tt = update(T[i-1],mp[a[i]],-1);
T[i] = update(tt,i,1);
}
mp[a[i]] = i;
}

int q;
scanf("%d",&q);

while(q--)
{
int l,r;
scanf("%d%d",&l,&r);
printf("%d\n",query(T[r],l));
}
}
return 0;
}