洛谷 P1960 列队

题意简述

有一个n × m 的矩阵，第i行第j列元素编号为（i - 1）× m +j
每次将一个数取出，其他元素依次向左，向上填补空缺，最后将取出的数放入矩阵最后一格
求每次取出数的编号

题解

由于最后一列较为特殊，只有当取出的数位于最后一列，向上填补空缺时才有影响，所有特殊考虑
对于每一行的（m - 1）个元素 和 最后一列元素都维护一个splay
进行删除第k个元素 和 在末尾插入元素 两个操作

代码

#include <cstdio>
#include <cassert>
using namespace std;
typedef long long ll;
struct Node
{
    ll l, r;
    Node *c[2];
    int s;
    Node() {l = r = 0; s = 0; c[0] = c[1] = 0; }
    Node(ll x) {l = r = x; s = 1; c[0] = c[1] = 0; }
    Node(ll l, ll r) {this -> l = l; this -> r = r; s = r - l + 1; c[0] = c[1] = 0; }
    int get(bool b){return c[b] ? c[b] -> s : 0; }
    void upt(){s = get(0) + get(1) + (r - l + 1); }
    int cmp(int x){return (x > get(0) && x <= get(0) + r - l + 1) ? -1 : x > get(0); }
}*r[300010];
int n, m, q, x, y;
void rtt(Node* &o, bool b)
{
    Node* t = o -> c[b ^ 1];
    o -> c[b ^ 1] = t -> c[b];
    t -> c[b] = o;
    o -> upt();
    t -> upt();
    o = t;
}
void splay(Node* &o, int k)
{
    if (!o) return;
    int d1 = o -> cmp(k);
    if (d1 == -1) return;
    int xx = k - d1 * (o -> s - o -> get(1));
    int d2 = o -> c[d1] -> cmp(xx);
    if (d2 == -1) rtt(o, d1 ^ 1);
    else 
    {
        splay(o -> c[d1] -> c[d2], xx - d2 * (o -> c[d1] -> s - o -> c[d1] -> get(1)));
        if (d1 == d2) rtt(o, d2 ^ 1);
        else rtt(o -> c[d1], d2 ^ 1);
        rtt(o, d1 ^ 1);
    }
}
void ins(Node* &o, Node* x)
{
    if (!o) {o = x; return; }
    splay(o, o -> s);
    o -> c[1] = x;
    o -> upt();
}
Node* del(Node* &o, int k)
{
    splay(o, k);
    if (o -> l == o -> r)
    {
        Node* t = o;
        if (!o -> c[0]) o = o -> c[1];
        else
        {
            splay(o -> c[0], o -> get(0));
            o -> c[0] -> c[1] = o -> c[1];
            o = o -> c[0];
            o -> upt();
        }
        t -> s = 1;
        t -> c[0] = t -> c[1] = 0;
        return t;
    }
    ll xxx = o -> l + k - o -> get(0) - 1;
    Node* x = new Node(o -> l, xxx - 1);
    o -> l = xxx + 1;
    x -> c[0] = o -> c[0];
    o -> c[0] = x;
    x -> upt();
    o -> upt();
    return new Node(xxx);
}
int main()
{
    scanf("%d%d%d", &n, &m, &q);
    for (register int i = 1; i <= n; ++i)
    {
        r[i] = new Node((ll)(i - 1) * m + 1, (ll)i * m - 1);
        ins(r[0], new Node((ll)i * m));
    }
    for (register int i = 1; i <= q; ++i)
    {
        scanf("%d%d", &x, &y);
        Node* idx;
        if (y == m)
        {
            printf("%lld\n", (idx = del(r[0], x)) -> l);
            ins(r[0], idx);
        }
        else
        {
            ins(r[x], del(r[0], x));
            printf("%lld\n", (idx = del(r[x], y)) -> l);
            ins(r[0], idx);
        }
    }
}