题目

　　https://vjudge.net/problem/Gym-247729C

题意

　　给出 n，m 表示初始序列 1 ~ n 以及 m 个操作，每个操作是将 p 位置开始之后的 s 个数放至最前面。

题解

　　题意十分简单，可是做不粗来，蒟蒻的我。首先可以想到，将一段区间 [ l,  r ] 放至最前方可以通过翻转 [ 1, l - 1 ] , [ l, r ] , [ 1, r ] 三个步骤得到，这样就转化成了 splay 的经典用法。这也是第一次学到splay。

#include <bits/stdc++.h>
// #include <iostream>
// #include <cstring>
// #include <string>
// #include <algorithm>
// #include <cmath>
// #include <cstdio>
// #include <queue>
// #include <stack>
// #include <map>
// #include <bitset>
// #include <set>
// #include <vector>
// #include <iomanip>
#define ll long long
#define ull unsigned long long
#define met(a, b) memset(a, b, sizeof(a))
#define rep(i, a, b) for(int i = a; i <= b; ++i)
#define bep(i, a, b) for(int i = a; i >= b; --i)
#define lowbit(x) (x&(-x))
#define MID (l + r) / 2
#define ls pos*2
#define rs pos*2+1
#define pb push_back
#define ios() ios::sync_with_stdio(0), cin.tie(0), cout.tie(0)

using namespace std;

const int maxn = 1e6 + 1010;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3f;
const ll mod = 1e9 + 7;
const double eps = 1e-4;
const double PI = acos(-1);

struct Splay_tree {
    int val, fa, sub_size, now_size, tag;
    int son[2];
}tree[maxn];

int root;
int arr[maxn];
int cnt;
int if_;

void push_down(int x) {
    if(x && tree[x].tag) {
        tree[tree[x].son[0]].tag ^= 1;
        tree[tree[x].son[1]].tag ^= 1;
        tree[x].tag = 0;
        swap(tree[x].son[0], tree[x].son[1]);
    }
}
void update(int x) {
    if(x) {
        tree[x].sub_size = tree[x].now_size;
        tree[x].sub_size += tree[tree[x].son[0]].sub_size;
        tree[x].sub_size += tree[tree[x].son[1]].sub_size;
    }
}
int build_tree(int l, int r, int fa) {
    if(l > r) return 0;
    int mid = MID;
    int now = ++cnt;
    tree[now].val = arr[mid];
    tree[now].fa = fa;
    tree[now].now_size = 1;
    tree[now].tag = 0;
    tree[now].son[0] = build_tree(l, mid - 1, now);
    tree[now].son[1] = build_tree(mid + 1, r, now);
    update(now);
    return now;
}
void rotate(int x) {
    int fx = tree[x].fa, ffx = tree[fx].fa;
    int k = (tree[fx].son[1] == x);
    push_down(fx);
    push_down(x);
    tree[fx].son[k] = tree[x].son[k ^ 1];
    tree[tree[fx].son[k]].fa = fx;
    tree[x].son[k ^ 1] = fx;
    tree[fx].fa = x;
    tree[x].fa = ffx;
    if(ffx) tree[ffx].son[tree[ffx].son[1] == fx] = x;
    update(fx);
    update(x);
}
void splay(int x, int goal) {
    while(tree[x].fa != goal) {
        int y = tree[x].fa, z = tree[y].fa;
        if(z != goal) {
            ((tree[z].son[1] == y) == (tree[y].son[1] == x)) ? rotate(y) : rotate(x);
        }
        rotate(x);
    }
    if(!goal) root = x;
}
int find(int x) {
    int now = root;
    while(1) {
        push_down(now);
        if(x <= tree[tree[now].son[0]].sub_size) {
            now = tree[now].son[0];
        }
        else {
            x -= tree[tree[now].son[0]].sub_size + 1;
            if(!x) return now;
            now = tree[now].son[1];
        }
    }
}
void Reverse(int x, int y) {
    int l = find(x - 1);
    int r = find(y + 1);
    splay(l, 0);
    splay(r, l);
    int pos = tree[r].son[0];
    tree[pos].tag ^= 1;
}
void dfs(int x) {
    push_down(x);
    if(tree[x].son[0]) dfs(tree[x].son[0]);
    if(tree[x].val != inf && tree[x].val != -inf) {
        if(if_) cout << ' ';
        else if_ = 1;
        cout << tree[x].val;
    }
    if(tree[x].son[1]) dfs(tree[x].son[1]);
}

int main() {
    int n, m;
    cin >> n >> m;
    rep(i, 2, n + 1) arr[i] = i - 1;
    arr[1] = -inf;
    arr[n + 2] = inf;
    //前后加入数据是为了防止 reverse(1, n) 这种边界时 find 出错。
    root = build_tree(1, n + 2, 0);
    rep(i, 1, m) {
        int pos, len;
        cin >> pos >> len;
        int l = pos + 1, r = l + len - 1;
        Reverse(l, r);
        Reverse(2, l - 1);
        Reverse(2, r);
    }
    dfs(root);
    return 0;
}