BZOJ3678 wangxz与OJ

[]

It seems like a super "water problem" and it actually is. It's very proud to find that my code is the fastest and shortest up to now.

[Solution]

Just like the problem which was also called OJ in SCOI2014. Each node on the split-merge tree refers to a segment. This problem is much easier to code than that one of SCOI.

[Code]

Spent too much time on some small mistakes.

#include <cstdio>
#include <cctype>
#include <memory.h>
#include <algorithm>

using namespace std;

#ifdef WIN32
#define getRand() ((rand()<<16)|rand())
#else
#define getRand() (rand())
#endif

int nextInt() {
	int d, s = 0;
	bool flag = 0;
	do {
		d = getchar();
		if (d == '-')
			flag = 1;
	} while (!isdigit(d));
	do
		s = s * 10 + d - 48, d = getchar();
	while (isdigit(d));
	return flag ? -s : s;
}

typedef pair <int, int> npair;

const int maxn = 80009;

int n, m;
int rt, sz[maxn], ls[maxn], rs[maxn], vl[maxn], vr[maxn], w[maxn], tn;

inline int nodesz(const int& p) {
	return vr[p] - vl[p] + 1;
}

int newNode(int l0, int r0) {
	tn ++;
	vl[tn] = l0;
	vr[tn] = r0;
	sz[tn] = nodesz(tn);
	ls[tn] = 0;
	rs[tn] = 0;
	w[tn] = getRand();
	return tn;
}

inline void update(const int& p) {
	sz[p] = sz[ls[p]] + sz[rs[p]] + nodesz(p);
}

int merge(int p, int q) {
	if (!p)
		return q;
	else if (!q)
		return p;
	else if (w[p] > w[q]) {
		rs[p] = merge(rs[p], q);
		update(p);
		return p;
	}
	else {
		ls[q] = merge(p, ls[q]);
		update(q);
		return q;
	}
}

npair split(int p, int k) {
	if (!p)
		return npair(0, 0);
	else if (sz[ls[p]] == k) {
		int q = ls[p];
		ls[p] = 0;
		update(p);
		return npair(q, p);
	}
	else if (sz[ls[p]] + nodesz(p) == k) {
		npair r = npair(p, rs[p]);
		rs[p] = 0;
		update(p);
		return r;
	}
	else if (sz[ls[p]] < k && sz[ls[p]] + nodesz(p) > k) {
		int q = newNode(vl[p] + k - sz[ls[p]], vr[p]);
		vr[p] = vl[q] - 1;
		sz[p] = nodesz(p);
		rs[q] = rs[p];
		rs[p] = 0;
		update(p);
		update(q);
		return npair(p, q);
	}
	else if (sz[ls[p]] > k) {
		npair x = split(ls[p], k);
		ls[p] = x. second;
		update(p);
		return npair(x. first, p);
	}
	else {
		npair x = split(rs[p], k - sz[ls[p]] - nodesz(p));
		rs[p] = x. first;
		update(p);
		return npair(p, x. second);
	}
}

int kth(int p, int k) {
	if (k > sz[p])
		return -1;
	else if (k > sz[ls[p]] && k <= sz[ls[p]] + nodesz(p))
		return vl[p] + k - sz[ls[p]] - 1;
	else if (k <= sz[ls[p]])
		return kth(ls[p], k);
	else
		return kth(rs[p], k - sz[ls[p]] - nodesz(p));
}

void ptree(int p, bool flag = 0) {
	if (!p)
		return;
	ptree(ls[p], 1);
	if (vl[p] < vr[p])
		printf("%d ~ %d ", vl[p], vr[p]);
	else
		printf("%d ", vl[p]);
	ptree(rs[p], 1);
	if (!flag)
		putchar(10);
}

int main() {
#ifndef ONLINE_JUDGE
	freopen("in.txt", "r", stdin);
#endif

	srand(20338423);
	tn = 0;
	rt = 0;
	n = nextInt();
	m = nextInt();
	for (int i = 0; i < n; i ++) {
		int a = nextInt();
		rt = merge(rt, newNode(a, a));
	}
	while (m --) {
		int opt = nextInt();
		if (opt == 0) {
			int p = nextInt();
			int a = nextInt();
			int b = nextInt();
			npair x = split(rt, p);
			x. first = merge(x. first, newNode(a, b));
			rt = merge(x. first, x. second);
		}
		else if (opt == 1) {
			int a = nextInt();
			int b = nextInt();
			npair x = split(rt, b);
			npair y = split(x. first, a - 1);
			rt = merge(y. first, x. second);
		}
		else if (opt == 2) {
			int p = nextInt();
			printf("%d\n", kth(rt, p));
		}
	}
}