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[BZOJ1552][Cerc2007]robotic sort

[BZOJ1552][Cerc2007]robotic sort

试题描述

技术分享

输入

输入共两行,第一行为一个整数N,N表示物品的个数,1<=N<=100000。第二行为N个用空格隔开的正整数,表示N个物品最初排列的编号。

输出

输出共一行,N个用空格隔开的正整数P1,P2,P3…Pn,Pi表示第i次操作前第i小的物品所在的位置。 注意:如果第i次操作前,第i小的物品己经在正确的位置Pi上,我们将区间[Pi,Pi]反转(单个物品)。

输入示例

6
3 4 5 1 6 2

输出示例

4 6 4 5 6 6

数据规模及约定

见“输入

题解

暴力用 splay 模拟。。。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cctype>
#include <algorithm>
using namespace std;

int read() {
	int x = 0, f = 1; char c = getchar();
	while(!isdigit(c)){ if(c == ‘-‘) f = -1; c = getchar(); }
	while(isdigit(c)){ x = x * 10 + c - ‘0‘; c = getchar(); }
	return x * f;
}

#define maxn 100010
int A[maxn], num[maxn], tot[maxn];

struct Node {
	int siz; bool rev;
	Node() {}
} ns[maxn];
int ToT, fa[maxn], ch[maxn][2];
void maintain(int o) {
	ns[o].siz = 1;
	for(int i = 0; i < 2; i++) if(ch[o][i])
		ns[o].siz += ns[ch[o][i]].siz;
	return ;
}
void build(int& o, int l, int r) {
	if(l > r) return ;
	int mid = l + r >> 1; o = A[mid];
	build(ch[o][0], l, mid - 1); build(ch[o][1], mid + 1, r);
	if(ch[o][0]) fa[ch[o][0]] = o;
	if(ch[o][1]) fa[ch[o][1]] = o;
	return maintain(o);
}
void pushdown(int o) {
	for(int i = 0; i < 2; i++) if(ch[o][i])
		ns[ch[o][i]].rev ^= ns[o].rev;
	if(ns[o].rev) swap(ch[o][0], ch[o][1]), ns[o].rev = 0;
	return ;
}
void rotate(int u) {
	int y = fa[u], z = fa[y], l = 0, r = 1;
	if(z) ch[z][ch[z][1]==y] = u;
	if(ch[y][1] == u) swap(l, r);
	fa[u] = z; fa[y] = u; fa[ch[u][r]] = y;
	ch[y][l] = ch[u][r]; ch[u][r] = y;
	maintain(y); maintain(u);
	return ;
}
int S[maxn], top;
void splay(int u) {
	int t = u; S[top = 1] = t;
	while(fa[t]) t = fa[t], S[++top] = t;
	while(top) pushdown(S[top--]);
	while(fa[u]) {
		int y = fa[u], z = fa[y];
		if(z) {
			if(ch[y][0] == u ^ ch[z][0] == y) rotate(u);
			else rotate(y);
		}
		rotate(u);
	}
	return ;
}
int Find(int o, int k) {
	pushdown(o);
	if(!o) return 0;
	int ls = ch[o][0] ? ns[ch[o][0]].siz : 0;
	if(k == ls + 1) return o;
	if(k > ls + 1) return Find(ch[o][1], k - ls - 1);
	return Find(ch[o][0], k);
}
int split(int u) {
	if(!u) return splay(1), 1;
	splay(u);
	int tmp = ch[u][1];
	fa[tmp] = ch[u][1] = 0;
	return maintain(u), tmp;
}
int merge(int a, int b) {
	if(!a) return b;
	pushdown(a); while(ch[a][1]) a = ch[a][1], pushdown(a);
	splay(a);
	ch[a][1] = b; fa[b] = a;
	return maintain(a), a;
}

int main() {
	int n = read();
	for(int i = 1; i <= n; i++) num[i] = A[i] = read();
	sort(num + 1, num + n + 1);
	for(int i = 1; i <= n; i++) {
		A[i] = lower_bound(num + 1, num + n + 1, A[i]) - num;
		A[i] += tot[A[i]]++;
	}
	 
	int tmp; build(tmp, 1, n);
	for(int i = 1; i <= n; i++) {
		splay(i); printf("%d%c", (ch[i][0] ? ns[ch[i][0]].siz : 0) + 1, i < n ? ‘ ‘ : ‘\n‘);
		int lrt = Find(i, i - 1), mrt = i, rrt;
		split(lrt); rrt = split(mrt); ns[mrt].rev ^= 1;
		mrt = merge(lrt, mrt); merge(mrt, rrt);
	}
	 
	return 0;
}

 

[BZOJ1552][Cerc2007]robotic sort