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luogu 2617

时间:2018-09-04 13:37:50      阅读:183      评论:0      收藏:0      [点我收藏+]

标签:树查询   cstring   uniq   names   fine   位置   getc   har   区间   

动态区间 $k$ 大
主席树 + 树状数组
树状数组的每个点对应一颗线段树
首先将所有点加入数据结构
枚举 x
code: for(int i = x; i <= n; i += Lowbit(i)) Poi_G(root[i], 1, Length, k, val);
区间修改时
将所有的后缀树的相应位置 -1, 再 +1
主席树查询时
在计算区间和的时候
类似树状数组的查询
将用到的线段树的相应节点加或减

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#include <cstring>
#include <string>

using namespace std;

#define LL long long

#define gc getchar()
inline int read() {int x = 0; char c = gc; while(c < 0 || c > 9) c = gc;
while(c >= 0 && c <= 9) x = x * 10 + c - 0, c = gc; return x;}
inline LL read_LL() {LL x = 0; char c = gc; while(c < 0 || c > 9) c = gc;
while(c >= 0 && c <= 9) x = x * 10 + c - 0, c = gc; return x;}
#undef gc

const int N = 1e5 + 10;

struct Node {int opt, l, r, k;} Ask[N];

int n, m;
int A[N], Num[N << 1], Length;

int root[N], Lson[N * 400], Rson[N * 400], W[N * 400];
int root_add[30], root_cut[30];
int jsadd, jscut;

int Lowbit(int x) {return (x & (-x));}

int Hjt;

void Poi_G(int &rt, int l, int r, int k, int val) {
    if(!rt) rt = ++ Hjt;
    W[rt] += val;
    if(l == r) return ;
    int mid = (l + r) >> 1;
    if(k <= mid) Poi_G(Lson[rt], l, mid, k, val);
    else Poi_G(Rson[rt], mid + 1, r, k, val);
}

void Pre_Poi_G(int x, int val) {
    int k = lower_bound(Num + 1, Num + Length + 1, A[x]) - Num;
    for(int i = x; i <= n; i += Lowbit(i)) Poi_G(root[i], 1, Length, k, val);
}

int Sec_A(int l, int r, int k) {
    if(l == r) return l;
    int sum = 0;
    for(int i = 1; i <= jsadd; i ++) sum += W[Lson[root_add[i]]];
    for(int i = 1; i <= jscut; i ++) sum -= W[Lson[root_cut[i]]];
    int mid = (l + r) >> 1;
    if(k <= sum) {
        for(int i = 1; i <= jsadd; i ++) root_add[i] = Lson[root_add[i]];
        for(int i = 1; i <= jscut; i ++) root_cut[i] = Lson[root_cut[i]];
        return Sec_A(l, mid, k);
    } else {
        for(int i = 1; i <= jsadd; i ++) root_add[i] = Rson[root_add[i]];
        for(int i = 1; i <= jscut; i ++) root_cut[i] = Rson[root_cut[i]];
        return Sec_A(mid + 1, r, k - sum);    
    }
}

int Pre_Sec_A(int l, int r, int k) {
    memset(root_add, 0, sizeof root_add);
    memset(root_add, 0, sizeof root_add);
    jsadd = jscut = 0;
    for(int i = r; i; i -= Lowbit(i)) root_add[++ jsadd] = root[i];
    for(int i = l - 1; i; i -= Lowbit(i)) root_cut[++ jscut] = root[i];
    return Sec_A(1, Length, k);
}

int main() {
    n = read(), m = read();
    for(int i = 1; i <= n; i ++) A[i] = read(), Num[++ Length] = A[i];
    for(int i = 1; i <= m; i ++) {
        char c[2]; scanf("%s", c);
        Ask[i].opt = (c[0] == Q ? 1 : 0);
        if(Ask[i].opt == 1) Ask[i].l = read(), Ask[i].r = read(), Ask[i].k = read();
        else {Ask[i].l = read(), Ask[i].k = read(), Num[++ Length] = Ask[i].k;}
    }
    sort(Num + 1, Num + Length + 1);
    Length = unique(Num + 1, Num + Length + 1) - Num - 1;
    for(int i = 1; i <= n; i ++) Pre_Poi_G(i, 1);
    for(int i = 1; i <= m; i ++) {
        if(Ask[i].opt == 1) {
            int Ans = Num[Pre_Sec_A(Ask[i].l, Ask[i].r, Ask[i].k)];
            printf("%d\n", Ans);
        } else {
            Pre_Poi_G(Ask[i].l, -1);
            A[Ask[i].l] = Ask[i].k;
            Pre_Poi_G(Ask[i].l, 1);
        }
    }
    return 0;
}

 

luogu 2617

标签:树查询   cstring   uniq   names   fine   位置   getc   har   区间   

原文地址:https://www.cnblogs.com/shandongs1/p/9583470.html

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