SPOJ - DQUERY 主席樹求區間中不同數字的個數

傳送門：SPOJ - DQUERY

題目大意

給定一個序列，求然後有q次查詢求[L,R]區間中不同數字的個數

解題思路

主席樹模版類型題目，離散化，更新查詢

AC代碼

#include <cstdio>
#include <iostream>
#include <map>
using namespace std;
const int MX = 1e5+5;
const int N = MX*20;
int n,q,tot;
int A[MX];
int T[N],lson[N],rson[N],sum[N];

int build(int l,int r)
{
    int rt = tot++;
    sum[rt] = 0;
    int mid = (l+r)>>1;
    if(l!=r) {
        lson[rt] = build(l,mid);
        rson[rt] = build(mid+1,r);
    }
    return rt;
}
int update(int rt,int pos,int val)
{
    int nrt = tot++;
    int tmp = nrt;
    int l=1,r=n;
    sum[nrt] = sum[rt] + val;
    while(l<r) {
        int mid = (l+r)>>1;
        if(pos<=mid) {
            lson[nrt] = tot++;
            rson[nrt] = rson[rt];
            nrt = lson[nrt];
            rt =  lson[rt];
            r  = mid;
        } else {
            rson[nrt] = tot++;
            lson[nrt] = lson[rt];
            nrt = rson[nrt];
            rt = rson[rt];
            l = mid+1;
        }
        sum[nrt] = sum[rt]+val;
    }
    return tmp;
}
int query(int rt,int pos)
{
    int  l= 1,r = n,ret = 0;
    while(pos>l) {
        int mid = (l+r)>>1;
        if(pos>mid)  {
            l = mid+1;
            rt= rson[rt];
        } else{
            r  = mid;
            ret += sum[rson[rt]];
            rt  =  lson[rt];
        }
    }
    return ret+sum[rt];
}

int main()
{freopen("in.txt","r",stdin);
    scanf("%d",&n);
    for(int i=1;i<=n;i++) scanf("%d",&A[i]);
    T[0] = build(1,n);
    map<int,int>mp;
    for(int i=1;i<=n;i++) {
        if(mp.find(A[i])==mp.end()) T[i] =  update(T[i-1],i,1);
        else {
            int tmp  = update(T[i-1],mp[A[i]],-1);
            T[i]  = update(tmp,i,1);
        }
        mp[A[i]] = i;
    }
    scanf("%d",&q);
    while(q--)  {
        int l,r;
        scanf("%d%d",&l,&r);
        printf("%d\n",query(T[r],l));
    }
    return 0;
}