【题目大意】
一个$n$个数的序列，$m$次操作，每次选择一段区间$[l, r]$，求出$[l, r]$中出现超过一半的数。
如果没有超过一半的数，那么就把答案钦定为$s$，每次会有$k$个数进行改变，给出下标，改变成当前的答案$s$。
$n, m leq 5*10^5, sum kleq 10^6$
By FJSDFZ ditoly
【题解】
用这题的方法进行线段树操作即可：http://www.cnblogs.com/galaxies/p/20170602_c.html
但是这样需要验证是否可行，那么问题变成待修改的询问$[l, r]$区间内$x$的出现次数。
如果没有修改，我会分块！
修改就写个平衡树（开始想STL发现set不支持iterator减法，就gg了）
然后慢的要死。（论不会treap的危害）
后来一气之下找了pb_ds::tree来用。
# include <ext/pb_ds/assoc_container.hpp> # include <ext/pb_ds/tree_policy.hpp> using namespace __gnu_pbds; tree <int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> S[M];
每次只要调用order_of_key(x)即可，注意这个操作是右开区间，所以本题中可能需要些许变换。
# include <stdio.h> # include <string.h> # include <iostream> # include <algorithm> // # include <bits/stdc++.h> using namespace std; typedef long long ll; typedef long double ld; typedef unsigned long long ull; const int M = 5e5 + 10; const int mod = 1e9+7; inline int getint() { int x = 0; char ch = getchar(); while(!isdigit(ch)) ch = getchar(); while(isdigit(ch)) x = (x<<3) + (x<<1) + ch - '0', ch = getchar(); return x; } # define gi getint() int n, a[M]; struct node { int w, t; node() {} node(int w, int t) : w(w), t(t) {} friend node operator + (node a, node b) { if(a.w == b.w) return node(a.w, a.t + b.t); if(a.t >= b.t) return node(a.w, a.t - b.t); else return node(b.w, b.t - a.t); } }; /* int rt[M]; struct Splay { int ch[M][2], fa[M], val[M], s[M]; inline void up(int x) { if(!x) return ; s[x] = 1 + s[ch[x][0]] + s[ch[x][1]]; } inline void rotate(int x, int &rt) { int y = fa[x], z = fa[y], ls = ch[y][1] == x, rs = ls^1; if(rt == y) rt = x; else ch[z][ch[z][1] == y] = x; fa[ch[x][rs]] = y, fa[y] = x, fa[x] = z; ch[y][ls] = ch[x][rs]; ch[x][rs] = y; up(y), up(x); } inline void splay(int x, int &rt) { while(x != rt) { int y = fa[x], z = fa[y]; if(y != rt) { if((ch[z][0] == y) ^ (ch[y][0] == x)) rotate(x, rt); else rotate(y, rt); } rotate(x, rt); } } inline int find(int x, int pos) { if(pos == val[x]) return x; if(pos < val[x]) return find(ch[x][0], pos); else return find(ch[x][1], pos); } inline int gmax(int x) { while(ch[x][1]) x = ch[x][1]; return x; } inline int del(int &rt, int pos) { int x = find(rt, pos); splay(x, rt); if(!ch[x][0]) { rt = ch[x][1]; fa[rt] = 0; ch[x][1] = 0; return x; } if(!ch[x][1]) { rt = ch[x][0]; fa[rt] = 0; ch[x][0] = 0; return x; } int y = gmax(ch[x][0]); splay(y, ch[x][0]); rt = y; ch[y][1] = ch[x][1]; fa[ch[x][1]] = y; ch[x][0] = ch[x][1] = 0; fa[y] = 0; up(y); return x; } inline void ins(int &x, int y, int pos, int id) { if(!x) { x = id; fa[x] = y; val[x] = pos; s[x] = 1; ch[x][0] = ch[x][1] = 0; return ; } if(pos < val[x]) ins(ch[x][0], x, pos, id); else ins(ch[x][1], x, pos, id); up(x); } inline void Ins(int &rt, int pos, int id) { ins(rt, 0, pos, id); splay(id, rt); } inline int rank(int x, int d) { if(!x) return 0; if(d < val[x]) return rank(ch[x][0], d); else return s[ch[x][0]] + 1 + rank(ch[x][1], d); } inline void DEBUG(int x) { if(!x) return ; DEBUG(ch[x][0]); printf("x = %d, ls = %d, rs = %d, pos = %d, sz = %d ", x, ch[x][0], ch[x][1], val[x], s[x]); DEBUG(ch[x][1]); } inline void debug() { for (int i=1; i<=5; ++i) { printf("rt %d ====================== ", i); DEBUG(rt[i]); } puts(" "); } }S; */ # include <ext/pb_ds/assoc_container.hpp> # include <ext/pb_ds/tree_policy.hpp> using namespace __gnu_pbds; tree <int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> S[M]; struct SMT { node w[M << 2]; # define ls (x<<1) # define rs (x<<1|1) inline void up(int x) { w[x] = w[ls] + w[rs]; } inline void build(int x, int l, int r) { if(l == r) { w[x] = node(a[l], 1); return ; } int mid = l+r>>1; build(ls, l, mid), build(rs, mid+1, r); up(x); } inline void edt(int x, int l, int r, int ps, int d) { if(l == r) { S[w[x].w].erase(l); w[x] = node(d, 1); S[d].insert(l); return ; } int mid = l+r>>1; if(ps <= mid) edt(ls, l, mid, ps, d); else edt(rs, mid+1, r, ps, d); up(x); } inline node query(int x, int l, int r, int L, int R) { if(L <= l && r <= R) return w[x]; int mid = l+r>>1; if(R <= mid) return query(ls, l, mid, L, R); else if(L > mid) return query(rs, mid+1, r, L, R); else return query(ls, l, mid, L, mid) + query(rs, mid+1, r, mid+1, R); } inline void debug(int x, int l, int r) { printf("x = %d, [%d, %d], w[x] = {%d, %d} ", x, l, r, w[x].w, w[x].t); if(l == r) return ; int mid = l+r>>1; debug(ls, l, mid); debug(rs, mid+1, r); } # undef ls # undef rs }T; inline bool check(int l, int r, int d) { int times = S[d].order_of_key(r+1) - S[d].order_of_key(l); // cout << "times = " << times << endl; if(times <= (r-l+1)/2) return false; return true; } int main() { freopen("president.in", "r", stdin); freopen("president.out", "w", stdout); n = gi; int Q = gi; for (int i=1; i<=n; ++i) { a[i] = gi; S[a[i]].insert(i); } T.build(1, 1, n); int l, r, s, k, leader; node ans; while(Q--) { if(Q % 10000 == 0) cerr << Q << endl; // S.debug(); l = gi, r = gi, s = gi, k = gi; ans = T.query(1, 1, n, l, r); if(check(l, r, ans.w)) leader = ans.w; else leader = s; for (int i=1, t; i<=k; ++i) { t = gi; T.edt(1, 1, n, t, leader); } printf("%d ", leader); // T.debug(1, 1, n); } ans = T.query(1, 1, n, 1, n); if(check(1, n, ans.w)) printf("%d ", ans.w); else puts("-1"); return 0; }
注释里面是平衡树。。T得不要不要的。。
复杂度都是$O((n + m + sum K) log n)$