BZOJ 1503 郁闷的出纳员 Splay

这题由于每次都是对树全体进行操作,其实设置一个全局的延迟标记就好

用到的操作主要是插入,删除子树,统计子树信息找第k大。

感觉Splay可以灵活的把需要用到的节点提到根上或者是根的儿子这里还是很方便的。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cstdlib>
#include <climits>

using namespace std;

const int maxn = 1e6 + 10;

int ch[maxn][2], val[maxn], size[maxn], chcnt, fa[maxn], root;
char cmd[16];
int n, min_val, tmp, tval, lcnt;

void update(int x) {
	size[x] = 1;
	if (ch[x][0]) size[x] += size[ch[x][0]];
	if (ch[x][1]) size[x] += size[ch[x][1]];
}

void rotate(int x, int d) {
	int y = fa[x];
	fa[ch[x][d]] = y; ch[y][d ^ 1] = ch[x][d];
	if (fa[y]) ch[fa[y]][y == ch[fa[y]][1]] = x;
	fa[x] = fa[y]; fa[y] = x;
	ch[x][d] = y;
	update(x); update(y);
}

void splay(int x, int goal) {
	while (fa[x] != goal) {
		int y = fa[x], d = (x == ch[y][1]);
		if (fa[y] == goal) rotate(x, d ^ 1);
		else {
			int z = fa[y], d1 = (y == ch[z][1]);
			if (d == d1) {
				rotate(y, d ^ 1); rotate(x, d ^ 1);
			}
			else {
				rotate(x, d ^ 1); rotate(x, d1 ^ 1);
			}
		}
	}
	if (goal == 0) root = x;
}

int findkth(int x, int k) {
	if (k <= 0 || k > size[x] || x == 0) return -1;
	int rsize = (ch[x][1] == 0 ? 0 : size[ch[x][1]]);
	if (rsize >= k) return findkth(ch[x][1], k);
	else if (k == rsize + 1) return x;
	else return findkth(ch[x][0], k - 1 - rsize);
}

int newNode(int &r, int father, int v) {
	r = ++chcnt;
	ch[r][0] = ch[r][1] = 0;
	fa[r] = father; val[r] = v;
	size[r] = 1;
	return r;
}

int insert(int v) {
	int u = root;
	if (u == 0) {
		newNode(root, 0, v); return root;
	}
	while (ch[u][v > val[u]]) u = ch[u][v > val[u]];
	int r = newNode(ch[u][v > val[u]], u, v);
	update(u);
	splay(r, 0);
	return r;
}

void remove_tree(int x) {
	if (x == 0) return;
	if (x == root) root = 0;
	else {
		ch[fa[x]][x == ch[fa[x]][1]] = 0;
		x = fa[x]; fa[x] = 0;
		while (x) {
			update(x); x = fa[x];
		}
	}
}

void debug(int root) {
	int lc = ch[root][0], rc = ch[root][1];
	printf("当前节点:%d(val = %d), lch: %d(val = %d), rch %d(val = %d), fa is %d\n",
		root, val[root], lc, val[lc], rc, val[rc], fa[root]);
	if (lc) debug(lc);
	if (rc) debug(rc);
}

int main() {
	while (scanf("%d%d", &n, &min_val) != EOF) {
		chcnt = root = tmp = lcnt = 0;
		for (int i = 1; i <= n; i++) {
			scanf("%s%d", cmd, &tval);
			if (cmd[0] == 'I') {
				if (tval < min_val) continue;
				insert(tval - tmp);
			}
			else if (cmd[0] == 'A') tmp += tval;
			else if (cmd[0] == 'S') {
				tmp -= tval;
				insert(min_val - tmp);
				lcnt += size[ch[root][0]];
				remove_tree(ch[root][0]);
				fa[ch[root][1]] = 0;
				root = ch[root][1];
			}
			else {
				int ret = findkth(root, tval);
				printf("%d\n", ret == -1 ? -1 : val[ret] + tmp);
			}
		}
		printf("%d\n", lcnt);
	}
	return 0;
}

  

posted @ 2015-02-06 15:02  acm_roll  阅读(184)  评论(0编辑  收藏  举报