洛谷P3960 列队 Splay

其实思路并不算太难，就是代码量相对较大。
我们将一次离队转换为一次删除和两次加入，这样就可以保证空间是动态分配的，最大也不会暴空间。
说实话写之前感觉会很恶心，但是代码量就还好吧，有些细节需要特殊注意一下。
Code：

#include<cstdio>
using namespace std;
const int maxn = 2000000 + 5;
long long val[maxn], arr[maxn]; 
int f[maxn], ch[maxn][2], root[maxn], siz[maxn], nums[maxn], cnt;
struct Operator
{
    inline void pushup(int x) { siz[x] = siz[ch[x][0]] + nums[x] + siz[ch[x][1]]; }
    void build(int l,int r,int fa,int &cur)
    {
        if(l > r) return;
        cur = ++ cnt;
        int mid = (l + r) >> 1;
        f[cur] = fa, val[cur] = arr[mid], nums[cur] = 1; 
        build(l,mid - 1,cur,ch[cur][0]);
        build(mid + 1, r, cur, ch[cur][1]);
        pushup(cur);
    }
    inline int get(int x){return ch[f[x]][1] == x;}
    inline void rotate(int x)
    {
        int old = f[x], oldf = f[old], which = get(x);
        ch[old][which] = ch[x][which ^ 1], f[ch[old][which]] = old;
        ch[x][which ^ 1] = old, f[old] = x, f[x] = oldf;
        if(oldf)ch[oldf][ch[oldf][1] == old] = x;
        pushup(old);pushup(x);
    }
    inline void splay(int x,int & tar)
    {
        int a = f[tar];
        for(int fa; (fa = f[x]) != a; rotate(x))
            if(f[fa] != a) rotate(get(x) == get(fa) ? fa : x);
        tar = x;
    }
    inline void insert_x(long long  x,int ty,int pos,int num)
    {
        if(!root[ty])
        {
            root[ty] = ++cnt, val[root[ty]] = x, nums[root[ty]] = num;
            pushup(root[ty]); 
            return; 
        }
        int p = root[ty], fa, cur;
        while(p)
        {
            fa = p;
            if(pos >= siz[ch[p][0]] + nums[p]) pos -= siz[ch[p][0]] + nums[p], p = ch[p][1], cur = 1;
            else p = ch[p][0], cur = 0;
        }
        ++cnt;
        val[cnt] = x, nums[cnt] = num;
        ch[fa][cur] = cnt, f[cnt] = fa;
        pushup(cnt);
        splay(cnt,root[ty]);
    }
    inline int get_node(int ty,int pos)        
    {
        int p = root[ty];
        while(pos)
        {
            if(siz[ch[p][0]] + nums[p] < pos) { pos -= siz[ch[p][0]] + nums[p], p = ch[p][1]; continue;}
            else if(pos - siz[ch[p][0]] > 0) break;
            else p = ch[p][0];
        }
        return p;
    }
    inline void delete_x(int ty, int pos)
    {
        int p = get_node(ty,pos);
        splay(p,root[ty]);
        if(!ch[p][0] && !ch[p][1]) {root[ty] = 0; return; }
        if(!ch[p][0]){ root[ty] = ch[p][1], f[ch[p][1]] = 0; }
        else if(!ch[p][1]) { root[ty] = ch[p][0], f[ch[p][0]] = 0;}
        else
        {
            p = ch[p][0];
            while(ch[p][1]) p = ch[p][1];
            splay(p,ch[root[ty]][0]);
            ch[p][1] = ch[root[ty]][1], f[ch[p][1]] = p, f[p] = 0;
            pushup(p);
            root[ty] = p;
        }
    }
    inline long long query(int ty,int x)
    {
        int p = get_node(ty,x);
        splay(p,root[ty]);
        return val[root[ty]] + x - siz[ch[root[ty]][0]] - 1;
    }
}T;
int main()
{
    int n,m,q;
    long long k = 0;
    scanf("%d%d%d",&n,&m,&q);
    for(int i = 1;i <= n;++i)
    {
        T.insert_x(k + 1,i,0,m - 1);
        k += m; 
        arr[i] = k;
    }
    T.build(1,n,0,root[n + 1]);
    for(int i = 1;i <= q; ++i)
    {
        int x,y;
        scanf("%d%d",&x,&y);
        long long last = T.query(n + 1,x);
        T.delete_x(n + 1,x);
        if(y == m)
        {
            printf("%lld
",last);
            T.insert_x(last,n + 1,siz[root[n + 1]],1);
        }
        else
        {
            long long ans = T.query(x,y);
            printf("%lld
",ans);
            int l = ans - val[root[x]];
            int r = val[root[x]] + nums[root[x]] - 1 - ans;
            int st = siz[ch[root[x]][0]] ;
            T.delete_x(x,y);
            if(l > 0) T.insert_x(ans - l, x, st , l);
            if(r > 0) T.insert_x(ans + 1, x, st + l,r);
            T.insert_x(ans,n + 1,siz[root[n + 1]], 1);
            T.insert_x(last,x,siz[root[x]],1);
        }
    }
    return 0;
}