可持久化线段树

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#数据结构 #可持久化 #线段树 #主席树

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例题

查询区间第k大

Code POJ 3237

#include <cstdio>
#include <algorithm>
#include <iostream>
using namespace std;

const int maxn = 1e5 + 5;
const int inf = 1e9 + 1;

struct segtree{

    int tot, rt[maxn];
    struct node{
        int lson, rson, size;
    }nd[maxn*40];

    void insert(int &i, int left, int rght, int x){
        int j = ++tot;
        int mid = (left + rght) >> 1;
        nd[j] = nd[i];
        nd[j].size++;
        i = j;
        if(left == rght) return;
        if(x <= mid) insert(nd[j].lson, left, mid, x);
        else insert(nd[j].rson, mid + 1, rght, x);
    }

    int query(int i, int j, int left, int rght, int k){
        if(left == rght) return left;
        int mid = (left + rght) >> 1;
        if(nd[nd[j].lson].size - nd[nd[i].lson].size >= k) return query(nd[i].lson, nd[j].lson, left, mid, k);
        else return query(nd[i].rson, nd[j].rson, mid + 1, rght, k - (nd[nd[j].lson].size - nd[nd[i].lson].size));
    }
}st;

int n, m;
int a[maxn], b[maxn], rnk[maxn], mp[maxn];
bool cmp(int i, int j){return a[i] < a[j];}

int main(){

    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; ++i) scanf("%d", &a[i]);
    for(int i = 1; i <= n; ++i) rnk[i] = i;
    sort(rnk + 1, rnk + 1 + n, cmp);
    a[0] = inf;
    for(int i = 1, j = 0; i <= n; ++i){
        int k = rnk[i], kk = rnk[i-1];
        if(a[k] != a[kk]) b[k] = ++j;
        else b[k] = j;
        mp[b[k]] = a[k];
    }
    for(int i = 1; i <= n; ++i) st.insert(st.rt[i] = st.rt[i-1], 1, n, b[i]);
    for(int i = 1; i <= m; ++i){
        int x, y, k;
        scanf("%d%d%d", &x, &y, &k);
        printf("%d\n", mp[st.query(st.rt[x-1], st.rt[y], 1, n, k)]);
    }

    return 0;
}

统计区间数字种数

Solution

用一个 $last$ 数组记录 $array$ 中每个数字上次出现的位置，现在问题就可以转化为统计 $[left,rght]$ 中 $last$ 值小于 $left$ 的个数。
用 $T_i$ 记录 $last[i]$ 的贡献，即在 $T_i$ 的 $pos$ 处插入一个 $+1$ 标记，查询的时候在 $T_{rght}-T_{left-1}$ 这颗线段树上查询 $[0,left-1]$ 的标记和。

Code HDU 5790

#include <map>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <iostream>
using namespace std;

typedef unsigned long long ull;

const int maxn = 2e6 + 5;
const ull p = 131;

int n, m;
char str[maxn];
map<ull, int> lst;

struct segtree{

    int tot, rt[maxn];
    struct node{
        int lson, rson, sum;
    }nd[maxn];

    void init(){
        tot = 0;
        nd[0].lson = nd[0].rson = nd[0].sum = 0;
    }

    void insert(int &i, int left, int rght, int pos){
        int j = ++tot;
        nd[j] = nd[i];
        nd[j].sum++;
        i = j;
        if(left == rght) return;
        int mid = (left + rght) >> 1;
        if(pos <= mid) insert(nd[j].lson, left, mid, pos);
        else insert(nd[j].rson, mid + 1, rght, pos);
    }

    int query(int i, int j, int left, int rght, int x, int y){
        if(x <= left && rght <= y) return nd[j].sum - nd[i].sum;
        int mid = (left + rght) >> 1;
        int lsum = 0, rsum = 0;
        if(x <= mid) lsum = query(nd[i].lson, nd[j].lson, left, mid, x, y);
        if(y > mid) rsum = query(nd[i].rson, nd[j].rson, mid + 1, rght, x, y);
        return lsum + rsum;
    }
}st;

void add(char s[], int id){

    st.rt[id] = st.rt[id-1];
    ull h = 0;
    for(int i = 0; s[i]; ++i){
        h *= p;
        h += s[i];
        st.insert(st.rt[id], 0, n, lst[h]);
        lst[h] = id;
    }
}

void work(){

    st.init();
    lst.clear();

    for(int i = 1; i <= n; ++i){
        scanf("%s", str);
        add(str, i);
    }

    scanf("%d", &m);
    for(int i = 1, Z = 0; i <= m; ++i){
        int left, rght, L, R;
        scanf("%d%d", &L, &R);
        left = min((Z + L) % n, (Z + R) % n) + 1;
        rght = max((Z + L) % n, (Z + R) % n) + 1;
        printf("%d\n", Z = st.query(st.rt[left-1], st.rt[rght], 0, n, 0, left - 1));
    }
}

int main(){

    while(scanf("%d", &n) == 1) work();

    return 0;
}