【HDU 3333 Turing Tree】离线处理+线段树+莫队

HDU 3333

题目大意：给你一串长度为n(n <= 30000)的序列，编号从1开始，序列中每个元素最大不超过10^9。
接下来Q组询问，每组是一个区间，要你回答这个区间中不相同的数之和为多少。

这道题之前自己做一直不会 学了莫队以后套个线段树过了

就是把询问先离线 按照 r 的大小从小到大排序 然后每到一个下标 就看这个下标的那个数之前有没有被插入过 如果插入过 就更新为新的下标新的值
如果不是 则直接插入
然后用一个cnt记录当前下标需不需要询问
注意答案数组和询问数组要开1E5的 不然会一直WA

/*
    if you can't see the repay
    Why not just work step by step
    rubbish is relaxed
    to ljq
*/
#include <cstdio>
#include <cstring>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <stack>
#include <set>
#include <sstream>
#include <vector>
#include <stdlib.h>
#include <algorithm>
using namespace std;

#define dbg(x) cout<<#x<<" = "<< (x)<< endl
#define dbg2(x1,x2) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<endl
#define dbg3(x1,x2,x3) cout<<#x1<<" = "<<x1<<" "<<#x2<<" = "<<x2<<" "<<#x3<<" = "<<x3<<endl
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))

typedef pair<int,int> pll;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll _INF = 0xc0c0c0c0c0c0c0c0;
const ll mod =  (int)1e9+7;

ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll ksm(ll a,ll b,ll mod){int ans=1;while(b){if(b&1) ans=(ans*a)%mod;a=(a*a)%mod;b>>=1;}return ans;}
ll inv2(ll a,ll mod){return ksm(a,mod-2,mod);}
void exgcd(ll a,ll b,ll &x,ll &y,ll &d){if(!b) {d = a;x = 1;y=0;}else{exgcd(b,a%b,y,x,d);y-=x*(a/b);}}//printf("%lld*a + %lld*b = %lld\n", x, y, d);

const int MAX_N = 30025;
const int MAX_M = 100025;
namespace sgt
{
    #define mid ((l+r)>>1)
    ll s[MAX_N<<2];
    void up(int rt)
    {
        s[rt] = s[rt<<1] +s[rt<<1|1];
    }
    void build(int rt,int l,int r)
    {
        s[rt] = 0;
        if(l==r)
        {
            return ;
        }
        build(rt<<1,l,mid);
        build(rt<<1|1,mid+1,r);
    }
    void update(int rt,int l,int r,int x,int v)
    {
        if(l==r)
        {
            s[rt] = v;
            return;
        }
        if(x<=mid) update(rt<<1,l,mid,x,v);
        else update(rt<<1|1,mid+1,r,x,v);
        up(rt);
    }
    ll query(int rt,int l,int r,int x,int y)
    {
        if(x<=l&&r<=y)
        {
            return s[rt];
        }
        if(x>mid) return query(rt<<1|1,mid+1,r,x,y);
        else if(y<=mid) return query(rt<<1,l,mid,x,y);
        else return query(rt<<1,l,mid,x,y)+query(rt<<1|1,mid+1,r,x,y);
    }
    #undef mid
}
ll ans[MAX_M],arr[MAX_N];
int cnt[MAX_N];
struct Query
{
    int l,r,id;
    bool operator < (const Query other) const
    {
        return r < other.r;
    }
}Q[MAX_M];
map<ll ,int > mp;
int main()
{
    //ios::sync_with_stdio(false);
    //freopen("a.txt","r",stdin);
    //freopen("b.txt","w",stdout);
    int n,t,q;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        mp.clear();
        sgt::build(1,1,n);
        for(int i = 1;i<=n;++i) cnt[i] = 0;
        for(int i = 1;i<=n;++i) scanf("%lld",&arr[i]);
        scanf("%d",&q);
        for(int i = 1;i<=q;++i)
        {
            scanf("%d%d",&Q[i].l,&Q[i].r);
            Q[i].id = i;
            cnt[Q[i].r]++;
        }
        sort(Q+1,Q+1+q);
        int l = 1;
        for(int i = 1;i<=n;++i)
        {
            if(mp.find(arr[i])==mp.end()) mp[arr[i]] = i;
            else
            {
                sgt::update(1,1,n,mp[arr[i]],0);
                mp[arr[i]] = i;
            }
            sgt::update(1,1,n,i,arr[i]);
            while(cnt[i])
            {
                cnt[i]--;
                ans[Q[l].id] = sgt::query(1,1,n,Q[l].l,Q[l].r);
                l++;
            }
        }
        for(int i = 1;i<=q;++i)
            printf("%lld\n",ans[i]);
    }
    //fclose(stdin);
    //fclose(stdout);
    //cout << "time: " << (long long)clock() * 1000 / CLOCKS_PER_SEC << " ms" << endl;
    return 0;
}