HDU 5737 Differencia set + 主席树

Posted 2022-10-28 cjlhy

HDU 5737

很明显能发现数据是随机的。

我们用set去维护值一样的段， 因为随机所以set里面线段的大小减少得很快， 平均大概30个左右。

每次查询暴力处理每一段相同的在主席树上查找累加答案就可以了。

但是这个好像不是标算， 标算复杂度(n + m) * log(n) 并且数据不是随机也可以。

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N = 1e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = (int)1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T& a, S b) a += b; if(a >= mod) a -= mod;
template<class T, class S> inline void sub(T& a, S b) a -= b; if(a < 0) a += mod;
template<class T, class S> inline bool chkmax(T& a, S b) return a < b ? a = b, true : false;
template<class T, class S> inline bool chkmin(T& a, S b) return a > b ? a = b, true : false;

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

int n, m, A[N], B[N];

struct Line 
    int l, r, val;
    bool operator < (const Line &rhs) const 
        return r < rhs.r;
    
;

set<Line> S;

int change(int L, int R, int val) 
    while(1) 
        auto it = S.lower_bound(Line0, L, 0);
        if(it == S.end() || it->l > R) break;
        Line tmp = *it;
        S.erase(it);
        if(tmp.l < L) 
            S.insert(Linetmp.l, L - 1, tmp.val);
        
        if(tmp.r > R) 
            S.insert(LineR + 1, tmp.r, tmp.val);
        
    
    S.insert(LineL, R, val);



int Rt[N], hs[N], hs_cnt;
struct ChairManTree 
    int tree_cnt;
    struct Node 
        int sum, ls, rs;
     a[N * 20];
    void update(int p, int l, int r, int &x, int y) 
        x = ++tree_cnt;
        a[x] = a[y];
        a[x].sum++;
        if(l == r) return;
        int mid = l + r >> 1;
        if(p <= mid) update(p, l, mid, a[x].ls, a[y].ls);
        else update(p, mid + 1, r, a[x].rs, a[y].rs);
    
    int query(int val, int l, int r, int x, int y) 
        if(hs[r] <= val) return a[x].sum - a[y].sum;
        if(hs[l] > val) return 0;
        int mid = l + r >> 1;
        return query(val, l, mid, a[x].ls, a[y].ls) + query(val, mid + 1, r, a[x].rs, a[y].rs);
    
 CT;

int query(int L, int R) 
    int ql, qr, ans = 0;
    auto it = S.lower_bound(Line0, L, 0);
    while(it != S.end() && it->l <= R) 
        ql = max(it->l, L);
        qr = min(it->r, R);
        ans += CT.query(it->val, 1, hs_cnt, Rt[qr], Rt[ql - 1]);
        ++it;
    
    return ans;


int a, b, C = ~(1 << 31), M = (1 << 16) - 1;

int rnd(int last) 
    a = (36969 + (last >> 3)) * (a & M) + (a >> 16);
    b = (18000 + (last >> 3)) * (b & M) + (b >> 16);
    return (C & ((a << 16) + b)) % 1000000000;


void init() 
    S.clear();
    CT.tree_cnt = 0;
    hs_cnt = 0;


int main() 
    int T; scanf("%d", &T);
    while(T--) 
        scanf("%d%d%d%d", &n, &m, &a, &b);
        init();
        for(int i = 1; i <= n; i++) 
            scanf("%d", &A[i]);
            S.insert(Linei, i, A[i]);
        
        for(int i = 1; i <= n; i++) 
            scanf("%d", &B[i]);
            hs[++hs_cnt] = B[i];
        
        sort(hs + 1, hs + 1 + hs_cnt);
        hs_cnt = unique(hs + 1, hs + 1 + hs_cnt) - hs - 1;
        for(int i = 1; i <= n; i++) 
            B[i] = lower_bound(hs + 1, hs + 1 + hs_cnt, B[i]) - hs;
            CT.update(B[i], 1, hs_cnt, Rt[i], Rt[i - 1]);
        
        int ans = 0;
        int last = 0, l, r, x;
        for(int i = 1; i <= m; i++) 
            l = rnd(last) % n + 1;
            r = rnd(last) % n + 1;
            if(l > r) swap(l, r);
            x = rnd(last) + 1;
            if((l + r + x) & 1) 
                change(l, r, x);
            
            else 
                last = query(l, r);
                add(ans, 1LL * i * last % mod);
            
        
        printf("%d\n", ans);
    
    return 0;


/*
*/