3600: 没有人的算术

3600: 没有人的算术

Time Limit: 20 Sec  Memory Limit: 128 MB
Submit: 616  Solved: 280
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湖北省队互测 Week1

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可以发现，每时每刻新增的元素加入之前元素组成的集合中，新集合内的元素两两之间一定可以找出严格的大小关系考虑用某数据结构维护已知集合内的元素，支持O(logn)查询，插入之类考虑当前新增第k个元素，前k - 1个元素的大小关系已经能在这个数据结构里体现了不妨给每个点一个实数，这样之前元素的比较就可以直接通过实数比较进行实数如何选取？给予每个节点一个区间[L,R]，定义该节点的势能phi[x] = (L[x] + R[x]) / 2这样通过势能比较就行了，能够比较，能够查询，此题的大部分操作就已经解决询问的话，用个线段树维护维护就完成了最后，数据结构的选择，不妨使用替罪羊树，因为区间的限制，使得节点之间的关系要相对静态替罪羊树的话，就是选取一个定值alpha，对于某个节点，如果其较大的子树的大小超过该点size * alpha，则标记这个点，每次插入操作，将标记最浅的那个子树暴力重构成一棵完全二叉树

具体的证明可见CLJ2013年的论文
第一遍写的时候fa[]和rt没有维护好(重构可能改变的东西= =)人傻不能怪社会系列。。。

#include<iostream>#include<cstdio>#include<algorithm>#include<cmath>#include<cstring>#include<vector>#include<queue>#include<set>#include<map>#include<stack>#include<bitset>#include<ext/pb_ds/priority_queue.hpp>using namespace std;const int maxn = 1E5 + 10;const int maxm = 5E5 + 50;const int T = 4;typedef double DB;const DB alpha = 0.666;const DB eps = 1E-9;DB lp,rp,siz[maxm],phi[maxm],L[maxm],R[maxm];int n,m,cnt,rt,tp,p,Now,Ans,fa[maxm],id[maxn*T],pos[maxn*T],ch[maxm][2],s[maxm],lc[maxm],rc[maxm];int getint(){char ch = getchar(); int ret = 0;while (ch < '0' || '9' < ch) ch = getchar();while ('0' <= ch && ch <= '9')ret = ret*10 + ch - '0',ch = getchar();return ret;}int getcom(){char ch = getchar();while (ch != 'C' && ch != 'Q') ch = getchar();return ch == 'C' ? 1 : 2;}int cmp(const int &x,const int &y){if (lc[x] == lc[y] && rc[x] == rc[y]) return 2;if (lc[x] == lc[y]) return phi[rc[x]] < phi[rc[y]] ? 1 : 0;return phi[lc[x]] < phi[lc[y]] ? 1 : 0;}void maintain(int o){int ls = (o<<1),rs = (o<<1|1);int d = cmp(id[ls],id[rs]);if (d == 1) pos[o] = pos[rs],id[o] = id[rs];else pos[o] = pos[ls],id[o] = id[ls];}void Build(int o,int l,int r){if (l == r) {id[o] = 1; pos[o] = l; return;}int mid = (l + r) >> 1;Build(o<<1,l,mid); Build(o<<1|1,mid+1,r);maintain(o);}void Modify(int o,int l,int r,int k,int ID){if (l == r) {id[o] = ID; return;}int mid = (l + r) >> 1;if (k <= mid) Modify(o<<1,l,mid,k,ID);else Modify(o<<1|1,mid+1,r,k,ID);maintain(o);}int Get_id(int o,int l,int r,int k){if (l == r) return id[o]; int mid = (l + r) >> 1;if (k <= mid) return Get_id(o<<1,l,mid,k);else return Get_id(o<<1|1,mid+1,r,k);}void Get_pos(int o,int l,int r,int ql,int qr){if (ql <= l && r <= qr){int d = cmp(id[o],Now);if (d == 0) Now = id[o],Ans = pos[o];else if (d == 2) Ans = min(Ans,pos[o]);return;}int mid = (l + r) >> 1;if (ql <= mid) Get_pos(o<<1,l,mid,ql,qr);if (qr > mid) Get_pos(o<<1|1,mid+1,r,ql,qr);}int Query(int x,int now){if (!x) return x;int d = cmp(x,now);if (d == 2) return x;return Query(ch[x][d],now);}void Insert(int &x,int now,DB LP,DB RP){if (!x){x = now; phi[x] = (LP + RP) / 2.00;L[x] = LP; R[x] = RP; siz[x] = 1.00; return;}DB mid = (LP + RP) / 2.00; int d = cmp(x,now);if (!d) Insert(ch[x][d],now,L[x],mid);else Insert(ch[x][d],now,mid,R[x]);fa[ch[x][d]] = x; siz[x] = siz[ch[x][0]] + siz[ch[x][1]] + 1.00;if (siz[ch[x][d]] >= alpha * siz[x]) p = x,lp = L[x],rp = R[x];}void Dfs(int x){if (!x) return; Dfs(ch[x][0]);s[++tp] = x; Dfs(ch[x][1]);}int Build_tree(int l,int r,DB LP,DB RP){if (l > r) return 0;int mid = (l + r) >> 1;int k = s[mid]; phi[k] = (LP + RP) / 2.00;ch[k][0] = Build_tree(l,mid-1,LP,phi[k]);ch[k][1] = Build_tree(mid+1,r,phi[k],RP);siz[k] = 1.00; L[k] = LP; R[k] = RP;for (int i = 0; i < 2; i++)if (ch[k][i]) fa[ch[k][i]] = k,siz[k] += siz[ch[k][i]];return k;}void Rebuild(){tp = 0; Dfs(p);int q = fa[p],now = Build_tree(1,tp,lp,rp);if (!q) rt = now,fa[now] = 0;else ch[q][ch[q][1] == p] = now,fa[now] = q;}int main(){#ifdef DMCfreopen("DMC.txt","r",stdin);#endifn = getint(); m = getint(); rt = cnt = 1;R[rt] = 1E9; phi[rt] = (L[rt] + R[rt]) / 2.00; Build(1,1,n);while (m--){int typ = getcom(),l,r;l = getint(); r = getint();if (typ == 1){int k = getint(); ++cnt;lc[cnt] = Get_id(1,1,n,l);rc[cnt] = Get_id(1,1,n,r);int ret = Query(rt,cnt);if (ret) Modify(1,1,n,k,ret),--cnt;else{Modify(1,1,n,k,cnt);p = 0; Insert(rt,cnt,L[rt],R[rt]);if (p) Rebuild();}}else{Now = 1; Ans = maxn;Get_pos(1,1,n,l,r); printf("%d\n",Ans);}}return 0;}