可以直接用treap上大模拟...n+1个treap维护n行的前m-1个点和最后一列。
需要支持删除一个点或者一段区间，而空间并不支持存下所有的点的时候，可以用一个点代替一个区间，记录区间首项的值和区间长度，这样每次查询某个点x的时候就可以用x在某个点y代表的区间里的rank来得到x的值，然后把x删去的时候，就把y这个区间从$[l,r]$拆分成$[l,x-1]$和$[x+1,r]$，重新加入。
类似的题有NOI超级钢琴
#include<iostream> #include<cstring> #include<cstdlib> #include<cstdio> #include<algorithm> #define ll long long #define int long long #define lt tree[x].ls #define rt tree[x].rs using namespace std; const int maxn=6000010; struct poi{ll beg; int rnd, size, ls, rs, len;}tree[maxn]; int n, m, Q, x, y, tott, tmp, rk; ll ans; int root[maxn], cnt[maxn], Len[maxn]; inline void read(int &k) { int f=1; k=0; char c=getchar(); while(c<‘0‘ || c>‘9‘) c==‘-‘ && (f=-1), c=getchar(); while(c<=‘9‘ && c>=‘0‘) k=k*10+c-‘0‘, c=getchar(); k*=f; } inline void build(int &x, ll beg, int len) { tree[x=++tott].beg=beg; tree[x].size=1; tree[x].len=Len[x]=len; tree[x].rnd=rand()<<15|rand(); } inline void up(int x) { tree[x].size=tree[lt].size+tree[rt].size+1; tree[x].len=tree[lt].len+tree[rt].len+Len[x]; } void split(int x, int &l, int &r, int k) { if(!k) l=0, r=x; else if(tree[x].size==k) l=x, r=0; else if(tree[lt].size>=k) r=x, split(lt, l, lt, k), up(x); else l=x, split(rt, rt, r, k-tree[lt].size-1), up(x); } void rank(int x, int k) { if(tree[lt].len<k && tree[lt].len+Len[x]>=k) ans=k-tree[lt].len, rk+=tree[lt].size+1; else if(k<=tree[lt].len) rank(lt, k); else rk+=tree[lt].size+1, rank(rt, k-tree[lt].len-Len[x]); } void merge(int &x, int l, int r) { if(!l || !r) x=l+r; else if(tree[l].rnd<tree[r].rnd) x=l, merge(rt, rt, r), up(x); else x=r, merge(lt, l, lt), up(x); } inline void disc(int &Root, ll &poi, int pos) { int x, y; rk=0; rank(Root, pos); split(Root, Root, y, rk); if(rk-1) split(Root, Root, x, rk-1); else x=Root; if(ans-1) build(tmp, tree[x].beg, ans-1), merge(Root, Root, tmp); if(Len[x]-ans) build(tmp, tree[x].beg+ans*(Root==root[n+1]?m:1), Len[x]-ans), merge(Root, Root, tmp); merge(Root, Root, y); poi=x; } inline ll query(int posx, int posy) { if(posy!=m) { ll ANS, poi; disc(root[posx], poi, posy); ANS=tree[poi].beg+ans-1; build(tmp, ANS, 1); merge(root[n+1], root[n+1], tmp); disc(root[n+1], poi, posx); poi=tree[poi].beg+(ans-1)*m; build(tmp, poi, 1); merge(root[posx], root[posx], tmp); return ANS; } ll poi; disc(root[n+1], poi, posx); poi=tree[poi].beg+(ans-1)*m; build(tmp, poi, 1); merge(root[n+1], root[n+1], tmp); return poi; } #undef int int main() { srand(19260817); read(n); read(m); read(Q); for(int i=1;i<=n;i++) build(root[i], 1ll*1+(i-1)*m, m-1); build(root[n+1], m, n); for(int i=1;i<=Q;i++) read(x), read(y), printf("%lld\n", query(x, y)); }
