HYSBZ 1901 Dynamic Rankings 树状数组套主席树

ZOJ上面这题内存限制太严格，裸的树套树主席树搞法过不去，BZOJ上面这个放的比较松，可以过。

其实就是利用树状数组维护n颗主席树，然后利用前缀和性质求解第k大。

#include <cstdio>
#include <cstring>
#include <iostream>
#include <map>
#include <set>
#include <vector>
#include <string>
#include <queue>
#include <deque>
#include <bitset>
#include <list>
#include <cstdlib>
#include <climits>
#include <cmath>
#include <ctime>
#include <algorithm>
#include <stack>
#include <sstream>
#include <numeric>
#include <fstream>
#include <functional>

using namespace std;

#define MP make_pair
#define PB push_back
typedef long long LL;
typedef unsigned long long ULL;
typedef vector<int> VI;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;
const int INF = INT_MAX / 3;
const double eps = 1e-8;
const LL LINF = 1e17;
const double DINF = 1e60;
const int maxn = 5e4 + 10;
const int maxm = maxn * 67;
const int maxq = 1e4 + 10;

struct Query {
	char cmd;
	int l, r, k;
	Query(char cmd = 0, int l = 0, int r = 0, int k = 0):
		cmd(cmd), l(l), r(r), k(k) {}
};

int sumv[maxm], lc[maxm], rc[maxm], cnt, C[maxn];
int n, m, num[maxn + maxq], numcnt, a[maxn];
Query ask[maxq];

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

void init() {
	cnt = numcnt = 0;
	memset(C, 0, sizeof(C));
}

void build(int rt, int l, int r) {
	sumv[rt] = 0;
	if(l == r) return;
	int mid = (l + r) >> 1;
	lc[rt] = cnt++; rc[rt] = cnt++;
	build(lc[rt], l, mid);
	build(rc[rt], mid + 1, r);
}

int update(int prt, int pos, int Val) {
	int nrt = cnt++, ret = nrt;
	sumv[nrt] = sumv[prt] + Val;
	int l = 0, r = numcnt - 1;
	while(l < r) {
		int mid = (l + r) >> 1;
		if(pos <= mid) {
			lc[nrt] = cnt++; rc[nrt] = rc[prt];
			sumv[lc[nrt]] = sumv[lc[prt]] + Val;
			nrt = lc[nrt]; prt = lc[prt];
			r = mid;
		}
		else {
			rc[nrt] = cnt++; lc[nrt] = lc[prt];
			sumv[rc[nrt]] = sumv[rc[prt]] + Val;
			nrt = rc[nrt]; prt = rc[prt];
			l = mid + 1;
		}
	}
	return ret;
}

void gao(int tid, int pos, int Val) {
	while(tid <= n) {
		int ret = update(C[tid], pos, Val);
		C[tid] = ret;
		tid += lowbit(tid);
	}
}

int lrt[maxn], rrt[maxn];
int query(int ql, int qr, int k) {
	int lcnt = 0, rcnt = 0;
	ql--;
	while(ql > 0) {
		lrt[lcnt++] = C[ql];
		ql -= lowbit(ql);
	}
	while(qr > 0) {
		rrt[rcnt++] = C[qr];
		qr -= lowbit(qr);
	}
	int l = 0, r = numcnt - 1;
	while(l < r) {
		int mid = (l + r) >> 1, lsum = 0, rsum = 0;
		for(int i = 0; i < lcnt; i++) lsum += sumv[lc[lrt[i]]];
		for(int i = 0; i < rcnt; i++) rsum += sumv[lc[rrt[i]]];
		if(rsum - lsum >= k) {
			r = mid;
			for(int i = 0; i < lcnt; i++) lrt[i] = lc[lrt[i]];
			for(int i = 0; i < rcnt; i++) rrt[i] = lc[rrt[i]];
		}
		else {
			l = mid + 1; k -= (rsum - lsum);
			for(int i = 0; i < lcnt; i++) lrt[i] = rc[lrt[i]];
			for(int i = 0; i < rcnt; i++) rrt[i] = rc[rrt[i]];
		}
	}
	return l;
}

int getid(int Val) {
	return lower_bound(num, num + numcnt, Val) - num;
}

int main() {
	init();
	scanf("%d%d", &n, &m);
	for(int i = 1; i <= n; i++) {
		scanf("%d", &a[i]);
		num[numcnt++] = a[i];
	}
	char cmd[5];
	int l, r, k;
	for(int i = 1; i <= m; i++) {
		scanf("%s", cmd);
		if(cmd[0] == ‘Q‘) {
			scanf("%d%d%d", &l, &r, &k);
			ask[i] = Query(cmd[0], l, r, k);
		}
		else {
			scanf("%d%d", &l, &k);
			ask[i] = Query(cmd[0], l, 0, k);
			num[numcnt++] = k;
		}
	}
	sort(num, num + numcnt);
	numcnt = unique(num, num + numcnt) - num;
	build(0, 0, numcnt - 1);
	for(int i = 1; i <= n; i++) {
		gao(i, getid(a[i]), 1);
	}
	for(int i = 1; i <= m; i++) {
		if(ask[i].cmd == ‘Q‘) printf("%d\n", num[query(ask[i].l, ask[i].r, ask[i].k)]);
		else {
			int prev = getid(a[ask[i].l]), now = getid(ask[i].k);
			a[ask[i].l] = ask[i].k;
			gao(ask[i].l , prev, -1); gao(ask[i].l, now, 1);
		}
	}
	return 0;
}