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【模板】可持久化线段树 1(主席树)

题目背景

这是个非常经典的主席树入门题——静态区间第K小

数据已经过加强,请使用主席树。同时请注意常数优化

题目描述

如题,给定N个正整数构成的序列,将对于指定的闭区间查询其区间内的第K小值。

输入输出格式

输入格式:

第一行包含两个正整数N、M,分别表示序列的长度和查询的个数。

第二行包含N个正整数,表示这个序列各项的数字。

接下来M行每行包含三个整数【模板】可持久化线段树 1(主席树)_#include

输出格式:

输出包含k行,每行1个正整数,依次表示每一次查询的结果

输入输出样例

输入样例#1: 复制

5 5
25957 6405 15770 26287 26465
2 2 1
3 4 1
4 5 1
1 2 2
4 4 1

输出样例#1: 复制

6405
15770
26287
25957
26287


主席树入门;
注意要离散化

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
//#pragma GCC optimize(2)
//#include
//#pragma GCC optimize("O3")
using namespace std;
#define maxn 700005
#define inf 0x3f3f3f3f
#define INF 9999999999
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)

inline ll rd() {
ll x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == '-') f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}

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

/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
if (!b) {
x = 1; y = 0; return a;
}
ans = exgcd(b, a%b, x, y);
ll t = x; x = y; y = t - a / b * y;
return ans;
}
*/



ll qpow(ll a, ll b, ll c) {
ll ans = 1;
a = a % c;
while (b) {
if (b % 2)ans = ans * a%c;
b /= 2; a = a * a%c;
}
return ans;
}

int n, m, q;
int cnt = 0;
int a[maxn], b[maxn], T[maxn];
int sum[maxn << 5], lson[maxn << 5], rson[maxn << 5];

int build(int l, int r) {
int rt = ++cnt;
sum[rt] = 0;
if (l < r) {
int mid = (l + r) >> 1;
lson[rt] = build(l, mid); rson[rt] = build(mid + 1, r);
}
return rt;
}

int upd(int pre, int l, int r, int x) {
int rt = ++cnt;
lson[rt] = lson[pre]; rson[rt] = rson[pre];
sum[rt] = sum[pre] + 1;
if (l < r) {
int mid = (l + r) >> 1;
if (x <= mid)lson[rt] = upd(lson[pre], l, mid, x);
else rson[rt] = upd(rson[pre], mid + 1, r, x);
}
return rt;
}

int query(int u, int v, int l, int r, int k) {
if (l >= r)return l;
int x = sum[lson[v]] - sum[lson[u]];
int mid = (l + r) >> 1;
if (x >= k)return query(lson[u], lson[v], l, mid, k);
else return query(rson[u], rson[v], mid + 1, r, k - x);
}

int main(){
//ios::sync_with_stdio(0);
rdint(n); rdint(q);
for (int i = 1; i <= n; i++)rdint(a[i]), b[i] = a[i];
sort(b + 1, b + 1 + n);
m = unique(b + 1, b + 1 + n) - b - 1;
T[0] = build(1, m);
for (int i = 1; i <= n; i++) {
int t = lower_bound(b + 1, b + 1 + m, a[i]) - b;
T[i] = upd(T[i - 1], 1, m, t);
}
while (q--) {
int x, y, z; rdint(x); rdint(y); rdint(z);
int t = query(T[x - 1], T[y], 1, m, z);
cout << b[t] << endl;
}
return 0;
}

 

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