P2617 Dynamic Rankings (dynamic segment tree prescription weight + Fenwick tree)

Meaning of the questions: repair request with a small interval k

Solution: Use the Chairman of the tree in memory demand interval k hour use of the prefix and the idea since it is a prefix and then we can use the prefix and better at maintaining the Fenwick tree

　　 But to preserve the array weight information but i dot every one here is not the tree line weights with a version of it in fact is not the Chairman of the tree in front of every one and not a common junction point

　　 For the first time we modify the operation of deleting this point and then update the original information into an array of points on the tree may jump together look at the code and understand a little better

　　 This method is also very strong restriction must Offline

 

#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 2;

int n, m, len, cnt;
int a[100005];
int b[200005];
int sum[MAXN * 400];
int ls[MAXN * 400];
int rs[MAXN * 400];
int t[MAXN];
int temp[2][50];
int tot1, tot0;

struct node {
    char opt;
    int u, v, w;
}E[100005];

void add(int &o, int l, int r, int k, int v) {
    if(!o) o = ++cnt;
    sum[o] += v;
    int mid = l + r >> 1;
    if(l == r) return;

    if(k <= mid) add(ls[o], l, mid, k, v);
    else add(rs[o], mid + 1, r, k, v);
}

void update(int x, int pos, int v) {
    for(int i = x; i <= n; i += (i & -i)) add(t[i], 1, len, pos, v);
}

void prepare_query(int l, int r) {
    tot1 = tot0 = 0;
    for(int i = r; i >= 1; i -= (i & -i)) temp[1][++tot1] = t[i];
    for(int i = l; i >= 1; i -= (i & -i)) temp[0][++tot0] = t[i];
}

int query(int l, int r, int k) {
    if(l == r) return l;

    int mid = l + r >> 1;
    int res = 0;
    for(int i = 1; i <= tot1; i++) res += sum[ls[temp[1][i]]];
    for(int i = 1; i <= tot0; i++) res -= sum[ls[temp[0][i]]];
    if(res >= k) {
        for(int i = 1; i <= tot1; i++) temp[1][i] = ls[temp[1][i]];
        for(int i = 1; i <= tot0; i++) temp[0][i] = ls[temp[0][i]];
        return query(l, mid, k);
    } else {
        for(int i = 1; i <= tot1; i++) temp[1][i] = rs[temp[1][i]];
        for(int i = 1; i <= tot0; i++) temp[0][i] = rs[temp[0][i]];
        return query(mid + 1, r, k - res);
    }
}

char s[5];
int main() {
    cnt = 0;
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; i++) scanf("%d", &a[i]), b[i] = a[i];
    len = n;
    
    for(int i = 1; i <= m; i++) {
        scanf("%s", s);
        E[i].opt = s[0];
        if(E[i].opt == 'Q') scanf("%d%d%d", &E[i].u, &E[i].v, &E[i].w);
        else {
            scanf("%d%d", &E[i].u, &E[i].v);
            b[++len] = E[i].v;
        }
    }
    sort(b + 1, b + 1 + len);
    len = unique(b + 1, b + 1 + len) - b - 1;

    for(int i = 1; i <= n; i++) {
        int tt = lower_bound(b + 1, b + 1 + len, a[i]) - b;
        update(i, tt, 1);
    }

    for(int i = 1; i <= m; i++) {
        if(E[i].opt == 'Q') {
            prepare_query(E[i].u - 1, E[i].v);
            printf("%d\n", b[query(1, len, E[i].w)]);
        } else {
            int t1 = lower_bound(b + 1, b + 1 + len, a[E[i].u]) - b;
            update(E[i].u, t1, -1);
            a[E[i].u] = E[i].v;
            int t2 = lower_bound(b + 1, b + 1 + len, a[E[i].u]) - b;
            update(E[i].u, t2, 1);
        }
    }
    return 0;
}

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