Codeforces 1139F Dish Shopping 树状数组套平衡树 || 平衡树

Dish Shopping

将每个物品拆成p 和 s 再加上人排序。 然后问题就变成了， 对于一个线段(L - R)，

问有多少个(li, ri)满足  L >= li && R >= ri， 这个东西可以直接树状数组套平衡树维护。

但是这个题目有个特殊性，因为排好序之后不会存在 li > L && ri > R的点， 所以可以直接

用平衡树， 或者线段树去维护这个东西。

平板电视

#include<bits/stdc++.h>
#include <bits/extc++.h>
#define LL 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 ull unsigned long long

using namespace std;
using namespace __gnu_pbds;

const int N = 1e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 998244353;
const double eps = 1e-6;
const double PI = acos(-1);

int n, m, tot, ans[N * 3], hs[N], cnt;
int p[N], s[N], b[N], inc[N], pref[N];

struct event {
    int p, b, id, op, rp;
    bool operator < (const event& rhs) const {
        if(p == rhs.p) return op < rhs.op;
        else return p < rhs.p;
    }
} e[N * 3];

template <class T>
using Tree = tree<T, null_type, std::less<T>, rb_tree_tag,tree_order_statistics_node_update>;

struct Bit {
    Tree<PII> T[N];
    void add(int x, PII v) {
        for(int i = x; i <= cnt; i += i & -i)
            T[i].insert(v);
    }
    void del(int x, PII v) {
        for(int i = x; i <= cnt; i += i & -i)
            T[i].erase(v);
    }
    int sum(int x, int R) {
        int ans = 0;
        for(int i = x; i; i -= i & -i)
            ans += T[i].order_of_key(mk(R, INT_MAX));
        return ans;
    }
} bit;

int getPos(int x) {
    return upper_bound(hs + 1, hs + 1 + cnt, x) - hs - 1;
}

int main() {
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i++) scanf("%d", &p[i]);
    for(int i = 1; i <= n; i++) scanf("%d", &s[i]);
    for(int i = 1; i <= n; i++) scanf("%d", &b[i]);
    for(int i = 1; i <= n; i++) {
        e[++tot] = event{p[i], b[i], i, 0, p[i]};
        e[++tot] = event{s[i], b[i], i, 2, p[i]};
        hs[++cnt] = p[i] - b[i];
    }
    for(int i = 1; i <= m; i++) scanf("%d", &inc[i]);
    for(int i = 1; i <= m; i++) scanf("%d", &pref[i]);
    for(int i = 1; i <= m; i++) e[++tot] = event{inc[i], pref[i], n + i, 1, 0};

    sort(hs + 1, hs + 1 + cnt);
    cnt = unique(hs + 1, hs + 1 + cnt) - hs - 1;

    sort(e + 1, e + 1 + tot);
    for(int i = 1; i <= tot; i++) {
        int p = e[i].p, b = e[i].b, rp = e[i].rp;
        if(e[i].op == 0) {
            bit.add(getPos(rp - b), mk(rp + b, e[i].id));
        } else if(e[i].op == 1) {
            ans[e[i].id] = bit.sum(getPos(p - b), p + b);
        } else {
            bit.del(getPos(rp - b), mk(rp + b, e[i].id));
        }
    }
    for(int i = n + 1; i <= n + m; i++) printf("%d ", ans[i]);
    puts("");
    return 0;
 }

/*
*/

treap, 为啥我的treap好慢啊啊啊。

#include<bits/stdc++.h>
#define LL 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 ull unsigned long long

using namespace std;

const int N = 1e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 998244353;
const double eps = 1e-6;
const double PI = acos(-1);

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

int n, m, tot, ans[N * 3], hs[N], cnt;
int p[N], s[N], b[N], inc[N], pref[N];

struct event {
    int p, b, id, op, rp;
    bool operator < (const event& rhs) const {
        if(p == rhs.p) return op < rhs.op;
        else return p < rhs.p;
    }
} e[N * 3];

struct node {
    node* ch[2];
    int key, fix, sz, cnt;
    void update() {
        sz = ch[0]->sz + ch[1]->sz + cnt;
    }
} base[N * 20];

typedef node* P_node;
P_node len = base;

struct Treap {
    node nil;
    P_node root, null;
    Treap() {
        root = null = &nil;
        null->key = null->fix = inf;
        null->sz = null->cnt = 0;
        null->ch[0] = null->ch[1] = null;
    }
    P_node newnode(int tkey) {
        len->key = tkey;
        len->fix = rng();
        len->ch[0] = len->ch[1] = null;
        len->sz = len->cnt = 1;
        return len++;
    }
    void rot(P_node &p, int d) {
        P_node k = p->ch[d ^ 1];
        p->ch[d ^ 1] = k->ch[d];
        k->ch[d] = p;
        p->update();
        k->update();
        p = k;
    }
    void _Insert(P_node &p, int tkey) {
        if(p == null) {
            p = newnode(tkey);
        } else if(p->key == tkey) {
            p->cnt++;
        } else {
            int d = tkey > p->key;
            _Insert(p->ch[d], tkey);
            if(p->ch[d]->fix > p->fix) {
                rot(p, d ^ 1);
            }
        }
        p->update();
    }

    void _Delete(P_node &p, int tkey) {
        if(p == null) return;
        if(p->key == tkey) {
            if(p->cnt > 1) p->cnt--;
            else if(p->ch[0] == null) p = p->ch[1];
            else if(p->ch[1] == null) p = p->ch[0];
            else {
                int d = p->ch[0]->fix > p->ch[1]->fix;
                rot(p, d);
                _Delete(p->ch[d], tkey);
            }
        } else {
            _Delete(p->ch[tkey > p->key], tkey);
        }
        p->update();
    }
    int _Kth(P_node p, int k) {
        if(p == null || k < 1 || k > p->sz) return 0;
        if(k < p->ch[0]->sz + 1) return _Kth(p->ch[0], k);
        if(k > p->ch[0]->sz + p->cnt) return _Kth(p->ch[1], k - p->ch[0]->sz - p->cnt);
        return p->key;
    }
    int _Rank(P_node p, int tkey, int res) {
        if(p == null) return -1;
        if(p->key == tkey) return p->ch[0]->sz + res + 1;
        if(tkey < p->key) return _Rank(p->ch[0], tkey, res);
        return _Rank(p->ch[1], tkey, res + p->ch[0]->sz + p->cnt);
    }
    int _Pred(P_node p, int tkey){
        if(p == null) return -inf;
        if(tkey <= p->key) return _Pred(p->ch[0], tkey);
        return max(p->key, _Pred(p->ch[1], tkey));
    }
    int _Succ(P_node p, int tkey){
        if(p == null) return inf;
        if(tkey >= p->key) return _Succ(p->ch[1], tkey);
        return min(p->key, _Succ(p->ch[0], tkey));
    }
    int _Query(P_node p, int tkey) {
        if(p == null) return 0;
        if(p->key > tkey) return _Query(p->ch[0], tkey);
        else if(p->key < tkey) return p->cnt + p->ch[0]->sz + _Query(p->ch[1], tkey);
        else return p->cnt + p->ch[0]->sz;
    }
    void Insert(int tkey){ _Insert(root,tkey); }
    void Delete(int tkey){ _Delete(root,tkey); }
    int Kth(int k){ return _Kth(root,k); }
    int Rank(int tkey){ return _Rank(root,tkey,0); }
    int Pred(int tkey){ return _Pred(root,tkey); }
    int Succ(int tkey){ return _Succ(root,tkey); }
    int Query(int tkey){ return _Query(root, tkey); }
}tp;

struct Bit {
    Treap T[N];
    void add(int x, int v) {
        for(int i = x; i <= cnt; i += i & -i)
            T[i].Insert(v);
    }
    void del(int x, int v) {
        for(int i = x; i <= cnt; i += i & -i)
            T[i].Delete(v);
    }
    int sum(int x, int R) {
        int ans = 0;
        for(int i = x; i; i -= i & -i)
            ans += T[i].Query(R);
        return ans;
    }
} bit;

int getPos(int x) {
    return upper_bound(hs + 1, hs + 1 + cnt, x) - hs - 1;
}

int main() {
    srand(time(NULL));
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i++) scanf("%d", &p[i]);
    for(int i = 1; i <= n; i++) scanf("%d", &s[i]);
    for(int i = 1; i <= n; i++) scanf("%d", &b[i]);
    for(int i = 1; i <= n; i++) {
        e[++tot] = event{p[i], b[i], i, 0, p[i]};
        e[++tot] = event{s[i], b[i], i, 2, p[i]};
        hs[++cnt] = p[i] - b[i];
    }
    for(int i = 1; i <= m; i++) scanf("%d", &inc[i]);
    for(int i = 1; i <= m; i++) scanf("%d", &pref[i]);
    for(int i = 1; i <= m; i++) e[++tot] = event{inc[i], pref[i], n + i, 1, 0};

    sort(hs + 1, hs + 1 + cnt);
    cnt = unique(hs + 1, hs + 1 + cnt) - hs - 1;

    sort(e + 1, e + 1 + tot);
    for(int i = 1; i <= tot; i++) {
        int p = e[i].p, b = e[i].b, rp = e[i].rp;
        if(e[i].op == 0) {
            bit.add(getPos(rp - b), rp + b);
        } else if(e[i].op == 1) {
            ans[e[i].id] = bit.sum(getPos(p - b), p + b);
        } else {
            bit.del(getPos(rp - b), rp + b);
        }
    }
    for(int i = n + 1; i <= n + m; i++) printf("%d ", ans[i]);
    puts("");
    return 0;
 }

/*
*/

原文地址：https://www.cnblogs.com/CJLHY/p/10612071.html