題面

SPOJ3267

Sol

給定\(N\)個正整數構成的序列，將對於指定的閉區間查詢其區間內的不同的數的個數

主席樹
不是權值線段樹
維護位置
如果插入一個數時發現之前有過了
那麽修改當前的，那個位置\(-1\)
然後插入這個數字，在相應的位置\(+1\)

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(3e4
 
 + 5);
const int __(1e6);

IL int Input(){
    RG int x = 0, z = 1; RG char c = getchar();
    for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
    for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48
 
);
    return x * z;
}

int n, m, rt[_], tot, o[_], len, a[_], vis[_];
struct HJT{
    int ls, rs, sz;
} T[__];

IL void Modify(RG int &x, RG int l, RG int r, RG int p, RG int v){
    T[++tot] = T[x], T[x = tot].sz += v;
    if(l == r) return;
    RG int mid = (l + r) >> 1;
    if(p <= mid) Modify(T[x].ls, l, mid, p, v);
    else
 
 Modify(T[x].rs, mid + 1, r, p, v);
}

IL int Query(RG int x, RG int l, RG int r, RG int L, RG int R){
    if(!x) return 0;
    if(L <= l && R >= r) return T[x].sz;
    RG int mid = (l + r) >> 1, ret = 0;
    if(L <= mid) ret = Query(T[x].ls, l, mid, L, R);
    if(R > mid) ret += Query(T[x].rs, mid + 1, r, L, R);
    return ret;
}

int main(RG int argc, RG char* argv[]){
    n = Input();
    for(RG int i = 1; i <= n; ++i) o[i] = a[i] = Input();
    sort(o + 1, o + n + 1), len = unique(o + 1, o + n + 1) - o - 1;
    for(RG int i = 1; i <= n; ++i){
        a[i] = lower_bound(o + 1, o + len + 1, a[i]) - o;
        rt[i] = rt[i - 1];
        if(!vis[a[i]]) vis[a[i]] = i, Modify(rt[i], 1, n, i, 1);
        else{
            Modify(rt[i], 1, n, vis[a[i]], -1);
            vis[a[i]] = i;
            Modify(rt[i], 1, n, i, 1);
        }
    }
    m = Input();
    for(RG int i = 1; i <= m; ++i){
        RG int l = Input(), r = Input();
        printf("%d\n", Query(rt[r], 1, n, l, r));
    }
    return 0;
}

SPOJ3267：D-query