[ZJU 2112] Dynamic Rankings

[题目链接]

          https://www.lydsy.com/JudgeOnline/problem.php?id=1901

[算法]

        首先 ， 考虑没有修改操作

        不妨建立可持久化线段树 ， 第i棵树维护区间[1 , i]中的数的出现个数 ， 则可以通过在线段树上二分的方式求出答案

        那么 ， 若有修改操作 ， 我们不妨使用树状数组套可持久化线段树 

        树状数组中的第i个元素为一棵可持久化线段树 ， 代表区间[i , i - lowbit(i) + 1]每个数的出现次数

        询问时可以同样二分 ， 只需用最多log(N)棵线段树作差即可

        时间复杂度 : O(NlogN ^ 2)

[代码]

          

#include<bits/stdc++.h>
using namespace std;
#define MAXN 200010
#define MAXP 5000010
#define MAXLOG 20
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;

int n , m , len , L , R;
int a[MAXN] , root[MAXN] , val[MAXN] , x[MAXN] , y[MAXN] , l[MAXN] , r[MAXN] , k[MAXN] , ql[MAXLOG] , qr[MAXLOG];
char type[MAXN][5];

struct Presitent_Segment_Tree
{
        int sz;
        int sum[MAXP] , lson[MAXP] , rson[MAXP];
        Presitent_Segment_Tree()
        {
                sz = 0;
                memset(root , 0 , sizeof(root));
                memset(lson , 0 , sizeof(lson));
                memset(rson , 0 , sizeof(rson));
        }
        inline void modify(int &k , int old , int l , int r , int pos , int value)
        {
                k = ++sz;
                lson[k] = lson[old] , rson[k] = rson[old];
                sum[k] = sum[old] + value;
                if (l == r) return;
                int mid = (l + r) >> 1;
                if (mid >= pos) modify(lson[k] , lson[k] , l , mid , pos , value);
                else modify(rson[k] , rson[k] , mid + 1 , r , pos , value);
        }
        inline int query(int l , int r , int k)
        {
                int cnt = 0;
                if (l == r)
                        return l;
                for (int i = 1; i <= L; i++) cnt -= sum[lson[ql[i]]];
                for (int i = 1; i <= R; i++) cnt += sum[lson[qr[i]]];
                int mid = (l + r) >> 1;
                if (cnt >= k)
                {   
                        for (int i = 1; i <= L; i++) ql[i] = lson[ql[i]];
                        for (int i = 1; i <= R; i++) qr[i] = lson[qr[i]];
                        return query(l , mid , k);
                } else
                {
                        for (int i = 1; i <= L; i++) ql[i] = rson[ql[i]];
                        for (int i = 1; i <= R; i++) qr[i] = rson[qr[i]];
                        return query(mid + 1 , r , k - cnt);
                }
        }
} PST;
struct Binary_Indexed_Tree
{
        inline int lowbit(int x)
        {
                return x & (-x);
        }
        inline void modify(int pos , int x , int val)
        {
                for (int i = pos; i <= n; i += lowbit(i))
                        PST.modify(root[i] , root[i] , 1 , len , x , val);
        }        
        inline int query(int l , int r , int k)
        {
                L = 0 , R = 0;
                for (int i = l - 1; i; i -= lowbit(i))
                        ql[++L] = root[i];
                for (int i = r; i; i -= lowbit(i))
                        qr[++R] = root[i];
                return PST.query(1 , len , k);
        }
} BIT;

template <typename T> inline void chkmax(T &x , T y) { x = max(x , y); }
template <typename T> inline void chkmin(T &x , T y) { x = min(x , y); }
template <typename T> inline void read(T &x)
{
        T f = 1; x = 0;
        char c = getchar();
        for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
        for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
        x *= f;
}

int main()
{

        scanf("%d%d" , &n , &m);
        len = n;
        for (int i = 1; i <= n; i++) 
        {
                scanf("%d" , &a[i]);
                val[i] = a[i];
        }
        for (int i = 1; i <= m; i++)
        {
                scanf("%s" , type[i]);
                if (type[i][0] == 'C')    
                {
                        scanf("%d%d" , &x[i] , &y[i]);
                        val[++len] = y[i];
                } else 
                        scanf("%d%d%d" , &l[i] , &r[i] , &k[i]);
        }
        sort(val + 1 , val + len + 1);
        len = unique(val + 1 , val + len + 1) - val - 1;
        for (int i = 1; i <= n; i++) a[i] = lower_bound(val + 1 , val + len + 1 , a[i]) - val;
        for (int i = 1; i <= m; i++)
                if (type[i][0] == 'C') y[i] = lower_bound(val + 1 , val + len + 1 , y[i]) - val;
        for (int i = 1; i <= n; i++) BIT.modify(i , a[i] , 1);
        for (int i = 1; i <= m; i++)
        {
                if (type[i][0] == 'C')
                {
                        BIT.modify(x[i] , a[x[i]] , -1);
                        BIT.modify(x[i] , y[i] , 1);
                        a[x[i]] = y[i];
                } else printf("%d
" , val[BIT.query(l[i] , r[i] , k[i])]);
        }
        
        return 0;
}