主席樹板子題

K - K-th Number
POJ - 2104
You are working for Macrohard company in data structures department. After failing your previous task about key insertion you were asked to write a new data structure that would be able to return quickly k-th order statistics in the array segment.
That is, given an array a[1...n] of different integer numbers, your program must answer a series of questions Q(i, j, k) in the form: "What would be the k-th number in a[i...j] segment, if this segment was sorted?"
For example, consider the array a = (1, 5, 2, 6, 3, 7, 4). Let the question be Q(2, 5, 3). The segment a[2...5] is (5, 2, 6, 3). If we sort this segment, we get (2, 3, 5, 6), the third number is 5, and therefore the answer to the question is 5.
Input

The first line of the input file contains n --- the size of the array, and m --- the number of questions to answer (1 <= n <= 100 000, 1 <= m <= 5 000).
The second line contains n different integer numbers not exceeding 109 by their absolute values --- the array for which the answers should be given.
The following m lines contain question descriptions, each description consists of three numbers: i, j, and k (1 <= i <= j <= n, 1 <= k <= j - i + 1) and represents the question Q(i, j, k).
Output

For each question output the answer to it --- the k-th number in sorted a[i...j] segment.
Sample Input

7 3
1 5 2 6 3 7 4
2 5 3
4 4 1
1 7 3
Sample Output

5
6
3
Hint

This problem has huge input,so please use c-style input(scanf,printf),or you may got time limit exceed.
題意：給你一個數組，告訴你若干個區間和k，讓你求出在這個區間裏升序排第k的數。
注意：
這題是主席樹裸的板子題，通常給你的空間上限是1e5或者常數級倍數由於數據量要開20倍。通常都須要進行離散化。只根據板子來的話主要操做爲4步，詳細在main函數的註釋裏。

 

板子一：

#include <string.h>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <deque>
#include <fstream>
#include <iomanip>
#include <iostream>
#include <iterator>
#include <list>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <stdexcept>
#include <string>
#include <vector>
using namespace std;
typedef unsigned long long ull;
#define ll long long
#define int long long
const int maxn = 1e5 + 10;
const int inf = 0x3f3f3f3f;
const int Base = 131;
const ll INF = 1ll << 62;
// const double PI = acos(-1);
const double eps = 1e-7;
const int mod = 999;
#define mem(a, b) memset(a, b, sizeof(a))
#define speed                        \
    {                                \
        ios::sync_with_stdio(false); \
        cin.tie(0);                  \
        cout.tie(0);                 \
    }

// inline int gcd(int a, int b) {
//     while (b ^= a ^= b ^= a %= b);
//     return a;
// }

inline ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a % b); }

long long fastPower(long long base, long long power)
{
    long long result = 1;
    while (power > 0)
    {
        if (power & 1)
            result = result * base % mod;
        power >>= 1;
        base = (base * base) % mod;
    }
    return result;
}

inline ll rd()
{
    ll s = 0, w = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9')
    {
        if (ch == '-')
            w = -1;
        ch = getchar();
    }
    while (ch >= '0' && ch <= '9')
        s = s * 10 + ch - '0', ch = getchar();
    return s * w;
}

int a[maxn];
int rt[maxn]; //rt[i]表示由數組前i個元素組成的線段樹的根結點
struct node
{
    int l, r; //線段樹左右子結點點
    int sum;  //結點信息，表示這顆子樹存在的元素的數目
} T[maxn * 20];
int tot = 0; //結點編號
vector<int> v;
int getid(int k)
{
    return lower_bound(v.begin(), v.end(), k) - v.begin() + 1;
}
void build(int &o, int l, int r) //創建一顆空樹
{
    o = ++tot;
    T[o].sum = 0;
    if (l == r)
        return;
    int mid = (l + r) / 2;
    build(T[o].l, l, mid);
    build(T[o].r, mid + 1, r);
}
void update(int l, int r, int &now, int last, int k)
{
    T[++tot] = T[last]; //複製線段樹(加入新的點)
    //更新當前線段樹的根結點
    now = tot;
    T[tot].sum++;
    if (l == r) //修改到葉子結點爲止
        return;
    //根據須要修改的k來肯定是修改左子樹仍是修改右子樹
    int mid = (l + r) >> 1;
    if (k <= mid)
        update(l, mid, T[now].l, T[last].l, k);
    else
        update(mid + 1, r, T[now].r, T[last].r, k);
}
int query(int l, int r, int x, int y, int k) //查詢區間【x，y】中第小的數
{
    if (l == r)
        return l; //查詢到葉子結點爲止
    int mid = (l + r) >> 1;
    int cnt = T[T[y].l].sum - T[T[x].l].sum; //第y顆樹比第x顆樹在左子樹上多的結點數
    if (cnt >= k)                             //答案在左子樹上
        return query(l, mid, T[x].l, T[y].l, k);
    else
        return query(mid + 1, r, T[x].r, T[y].r, k - cnt);
}

signed main()
{
    int n = rd();
    int m = rd();
    for (int i = 1; i <= n; i++)
    {
        a[i] = rd();
        v.push_back(a[i]);
    }
    sort(v.begin(), v.end());
    v.erase(unique(v.begin(), v.end()), v.end()); // 1.進行離散化注意
    build(rt[0], 1, n);                              //2.建樹注意
    for (int i = 1; i <= n; i++)                  //3.更新注意
        update(1, n, rt[i], rt[i - 1], getid(a[i]));
    while (m--)
    {
        int l = rd();
        int r = rd();
        int k = rd();
        //若是想要從大到小第k個那麼能夠令 k = (r - l + 1) - k + 1
        printf("%lld\n", v[query(1, n, rt[l - 1], rt[r], k) - 1]); //4.輸出注意
    }
    system("pause");
    return 0;
}

 

板子二：

#include <list>
#include <string.h>
#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <string>
#include <cstring>
#include <vector>
#include <map>
#include <deque>
#include <stack>
#include <queue>
#include <set>
#include <iomanip>
#include <cstdlib>
#include <stdexcept>
#include <fstream>
#include <iterator>
using namespace std;
typedef long long ll;
//const int maxn = 1e5+10;
const int inf = 0x3f3f3f3f;
const ll INF = 1ll << 62;
//const double PI = acos(-1);
const double eps = 1e-7;
const int mod = 998244353;
#define speed                        \
    {                                \
        ios::sync_with_stdio(false); \
        cin.tie(0);                  \
        cout.tie(0);                 \
    }
using namespace std;

const int MAXN = 1e5 + 5;

typedef struct node
{
    int x;
    int id;
} node;

int a[MAXN];
int ran[MAXN];
int rt[MAXN * 20], ls[MAXN * 20], rs[MAXN * 20], tot, size;
int sum[MAXN * 20];

void build(int &root, int l, int r)
{
    root = ++tot;
    sum[root] = 0;
    if (l == r)
        return;
    int mid = (l + r) >> 1;
    build(ls[root], l, mid);
    build(rs[root], mid + 1, r);
}

void update(int &root, int l, int r, int last, int x)
{
    root = ++tot;
    ls[root] = ls[last];
    rs[root] = rs[last];
    sum[root] = sum[last] + 1;
    if (l == r)
        return;
    int mid = (l + r) >> 1;
    if (mid >= x)
        update(ls[root], l, mid, ls[last], x);
    else
        update(rs[root], mid + 1, r, rs[last], x);
}

int query(int ss, int tt, int l, int r, int k)
{
    if (l == r)
        return l;
    int cnt = sum[ls[tt]] - sum[ls[ss]];
    int mid = (l + r) >> 1;
    if (k <= cnt)
        return query(ls[ss], ls[tt], l, mid, k);
    else
        return query(rs[ss], rs[tt], mid + 1, r, k - cnt);
}

int main()
{
    int n, m;
    scanf("%d %d", &n, &m);
    for (int i = 1; i <= n; i++)
    {
        scanf("%d", &a[i]);
        ran[i] = a[i];
    }
    tot = 0;
    sort(ran + 1, ran + n + 1);
    size = unique(ran + 1, ran + n + 1) - (ran + 1);
    build(rt[0], 1, size);
    for (int i = 1; i <= n; i++)
        a[i] = lower_bound(ran + 1, ran + 1 + size, a[i]) - ran;
    for (int i = 1; i <= n; i++)
        update(rt[i], 1, size, rt[i - 1], a[i]);
    while (m--)
    {
        int l, r, k;
        scanf("%d %d %d", &l, &r, &k);
        int id = query(rt[l - 1], rt[r], 1, size, k);
        //若是想要的是從大到小第k個能夠令 id = (r - l + 1) - id + 1
        printf("%d\n", ran[id]);
    }
    system("pause");
    return 0;
}
/*
1 
10 5 
1 4 2 3 5 6 7 8 9 0 
1 3 2 
*/