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  • hdu 5412 CRB and Queries

    题目连接

    http://acm.hdu.edu.cn/showproblem.php?pid=5412  

    CRB and Queries

    Description

    There are $N$ boys in CodeLand.
    Boy i has his coding skill $A_{i}$.
    CRB wants to know who has the suitable coding skill.
    So you should treat the following two types of queries.
    Query 1: 1 $l v$
    The coding skill of Boy l has changed to $v$.
    Query 2: 2 l $r k$
    This is a report query which asks the $k-th$ smallest value of coding skill between Boy $l$ and Boy $r$(both inclusive).

    Input

    There are multiple test cases.
    The first line contains a single integer $N$.
    Next line contains $N$ space separated integers $A_{1}, A_{2}, …, A_{N}$, where $A_{i}$ denotes initial coding skill of Boy $i$.
    Next line contains a single integer $Q$ representing the number of queries.
    Next $Q$ lines contain queries which can be any of the two types.
    $1 leq N, Q leq 10^{5} $
    $1 leq A_{i}, v leq 10^{9}$
    $1 leq l leq r leq N$
    $1 leq k leq r - l + 1$

    Output

    For each query of type 2, output a single integer corresponding to the answer in a single line.

    Sample Input

    5
    1 2 3 4 5
    3
    2 2 4 2
    1 3 6
    2 2 4 2

    Sample Output

    3
    4

    树套树裸题。。

    #include<algorithm>
    #include<iostream>
    #include<cstdlib>
    #include<cstring>
    #include<cstdio>
    #include<vector>
    #include<map>
    using std::map;
    using std::min;
    using std::find;
    using std::pair;
    using std::vector;
    using std::multimap;
    #define pb(e) push_back(e)
    #define sz(c) (int)(c).size()
    #define mp(a, b) make_pair(a, b)
    #define all(c) (c).begin(), (c).end()
    #define iter(c) __typeof((c).begin())
    #define cls(arr, val) memset(arr, val, sizeof(arr))
    #define cpresent(c, e) (find(all(c), (e)) != (c).end())
    #define rep(i, n) for(int i = 0; i < (int)n; i++)
    #define tr(c, i) for(iter(c) i = (c).begin(); i != (c).end(); ++i)
    #define lc root<<1
    #define rc root<<1|1
    const int N = 100010;
    const int INF = 0x3f3f3f3f;
    struct Node {
        int v, s, c;
        Node *ch[2];
        inline void push_up() {
            s = ch[0]->s + ch[1]->s + c;
        }
        inline void setc(int _v, int _s, Node *p) {
            v = _v, s = c = _s, ch[0] = ch[1] = p;
        }
        inline int cmp(int x) const {
            if (x == v) return -1;
            return x > v;
        }
    };
    int sum, arr[N];
    struct SBT {
        int top;
        Node *null, *tail, *pool[N], stack[N << 5], *ptr[N << 2];
        inline void init(int n) {
            top = 0;
            tail = &stack[0];
            null = tail++;
            null->setc(0, 0, NULL);
            for (int i = 1; i <= n; i++) scanf("%d", &arr[i]);
            seg_built(1, 1, n);
        }
        inline Node *newNode(int v) {
            Node *p = !top ? tail++ : pool[--top];
            p->setc(v, 1, null);
            return p;
        }
        inline void rotate(Node *&x, int d) {
            Node *k = x->ch[!d]; x->ch[!d] = k->ch[d]; k->ch[d] = x;
            k->s = x->s; x->push_up(); x = k;
        }
        inline void Maintain(Node *&x, int d) {
            if (!x->ch[d]->s) return;
            if (x->ch[d]->ch[d]->s > x->ch[!d]->s) rotate(x, !d);
            else if (x->ch[d]->ch[!d]->s > x->ch[!d]->s) rotate(x->ch[d], d), rotate(x, !d);
            else  return;
            Maintain(x, 0), Maintain(x, 1);
        }
        inline void insert(Node* &x, int key) {
            if (!x->s) { x = newNode(key); return; }
            int d = x->cmp(key);
            x->s++;
            if (-1 == d) { x->c++; return; }
            insert(x->ch[d], key);
            x->push_up();
            Maintain(x, d);
        }
        inline void erase(Node* &x, int key){
        if (!x->s) return;
        int d = x->cmp(key);
        x->s--;
        if (-1 == d) {
            if (x->c > 1) {
                x->c--;
            } else if (!x->ch[0]->s || !x->ch[1]->s) {
                pool[top++] = x;
                x = x->ch[0]->s ? x->ch[0] : x->ch[1];
            } else {
                Node *ret = x->ch[1];
                for (; ret->ch[0]->s; ret = ret->ch[0]);
                erase(x->ch[1], x->v = ret->v);
            }
        } else {
            erase(x->ch[d], key);
        }
        if (x->s) x->push_up();
    }
        inline int sbt_rank(Node *x, int key) {
            int t, cur = 0;
            for (; x->s;) {
                t = x->ch[0]->s;
                if (key < x->v) x = x->ch[0];
                else if (key >= x->v) cur += x->c + t, x = x->ch[1];
            }
            return cur;
        }
        inline void seg_built(int root, int l, int r) {
            ptr[root] = null;
            for (int i = l; i <= r; i++) insert(ptr[root], arr[i]);
            if (l == r) return;
            int mid = (l + r) >> 1;
            seg_built(lc, l, mid);
            seg_built(rc, mid + 1, r);
        }
        inline void seg_query(int root, int l, int r, int x, int y, int val) {
            if (x > r || y < l) return;
            if (x <= l && y >= r) {
                sum += sbt_rank(ptr[root], val);
                return;
            }
            int mid = (l + r) >> 1;
            seg_query(lc, l, mid, x, y, val);
            seg_query(rc, mid + 1, r, x, y, val);
        }
        inline void seg_modify(int root, int l, int r, int pos, int val) {
            if (pos > r || pos < l) return;
            erase(ptr[root], arr[pos]);
            insert(ptr[root], val);
            if (l == r) return;
            int mid = (l + r) >> 1;
            seg_modify(lc, l, mid, pos, val);
            seg_modify(rc, mid + 1, r, pos, val);
        }
        inline void kth(int n, int a, int b, int k) {
            int l = 0, r = INF;
            while (l < r) {
                sum = 0;
                int mid = (l + r) >> 1;
                seg_query(1, 1, n, a, b, mid);
                if (sum < k) l = mid + 1;
                else r = mid;
            }
            printf("%d
    ", l);
        }
        inline void solve(int n) {
            init(n);
            int m, a, b, c, k;
            scanf("%d", &m);
            while (m--) {
                scanf("%d", &a);
                if (2 == a) {
                    scanf("%d %d %d", &b, &c, &k);
                    kth(n, b, c, k);
                } else {
                    scanf("%d %d", &b, &c);
                    seg_modify(1, 1, n, b, c);
                    arr[b] = c;
                }
            }
        }
    }go;
    int main() {
    #ifdef LOCAL
        freopen("in.txt", "r", stdin);
        freopen("out.txt", "w+", stdout);
    #endif
        int n;
        while (~scanf("%d", &n)) {
            go.solve(n);
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/GadyPu/p/4796303.html
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