luogu P2617 Dynamic Rankings

传送门

动态区间第k小

想一下:

静态区间第k小 -> 主席树

动态整体第k小 -> 树状数组

所以树套树就行了

事实上这题还真就套一下 也不难写

一共两个操作:

1.插入 原来的一棵树插入变成树状数组插入

2.查询 原来的整体查询变成树状数组查询

修改就先-1再+1就行

可以开一个栈存每一组询问[l,r]中需要统计的主席树编号来统计答案

还是看代码吧 比较好理解

(主要是抄的时候挑了一个好的)

Code:

#include<cstdio>
#include<iostream>
#include<cstring>
#include<algorithm>
#define rep(i,a,n) for(int i = a;i <= n;i++)
#define per(i,n,a) for(int i = n;i >= a;i--)
using namespace std;
typedef long long ll;
int read() {
    int as = 0,fu = 1;
    char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') fu = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        as = as * 10 + c - '0';
        c = getchar();
    }
    return as * fu;
}
const int N = 400005;
const int M = 40000005;
//head

int n,Q,T;
int tot;
int sze[M];
int a[N],b[N],ca[N],cb[N],cc[N];
int ls[M],rs[M],rt[N];

void ins(int l,int r,int &t,int pre,int q,int v) {
    t = ++tot;
    sze[t] = sze[pre] + v;
    ls[t] = ls[pre],rs[t] = rs[pre];
    if(l == r) return;
    int m = l+r >> 1;
    if(q <= m) ins(l,m,ls[t],ls[pre],q,v);
    else ins(m+1,r,rs[t],rs[pre],q,v);
}

int tx[N],ty[N];
int query(int l,int r,int k) {
    if(l == r) return l;
    int sum = 0,m = l+r >> 1;
    rep(i,1,tx[0]) sum -= sze[ls[tx[i]]];
    rep(i,1,ty[0]) sum += sze[ls[ty[i]]];
    if(k <= sum) {
        rep(i,1,tx[0]) tx[i] = ls[tx[i]];
        rep(i,1,ty[0]) ty[i] = ls[ty[i]];
        return query(l,m,k);
    } else {
        rep(i,1,tx[0]) tx[i] = rs[tx[i]];
        rep(i,1,ty[0]) ty[i] = rs[ty[i]];
        return query(m+1,r,k - sum);
    }
}

inline void update(int x,int v) {
    int k = lower_bound(b + 1,b + T + 1,a[x]) - b;
    // printf("%d %d %d
",x,v,k);
    for(int i = x;i <= n;i += (i & (-i))) ins(1,T,rt[i],rt[i],k,v);
}

char cmd[20];
int main() {
    n = read(),Q = read();
    rep(i,1,n) a[i] = b[++T] = read();
    rep(i,1,Q) {
        scanf("%s",cmd);
        ca[i] = read(),cb[i] = read();
        cmd[0] == 'Q' ? cc[i] = read() : b[++T] = cb[i];
    }
    sort(b+1,b+T+1);
    T = unique(b+1,b+T+1) - b - 1;
    rep(i,1,n) update(i,1);
    rep(i,1,Q) {
        if(cc[i]) {
            tx[0] = ty[0] = 0;
            for(int j = ca[i]-1;j;j -= (j & (-j))) tx[++tx[0]] = rt[j];
            for(int j = cb[i];j;j -= (j & (-j))) ty[++ty[0]] = rt[j];
            printf("%d
",b[query(1,T,cc[i])]);
        } else update(ca[i],-1),a[ca[i]] = cb[i],update(ca[i],1);
    }
}