T1
给一个序列,求前 $k$ 大区间异或和的和
$n \leq 500000,k \leq min(n^2,200000)$
sol:
超级钢琴
对每个 $i$,维护一个三元组 $(l,r,i)$ 表示左端点在 $[l,r]$,右端点在 $i$ 的区间异或最值,维护一个堆,按这个异或最值排序,每次将堆顶拿出来,分裂成最多两个区间,查一下异或最大值即可,区间异或最大值可以前缀和+可持久化 Trie 来解决
#include <bits/stdc++.h> #define LL long long #define rep(i, s, t) for(register int i = (s), i##end = (t); i <= i##end; ++i) #define dwn(i, s, t) for(register int i = (s), i##end = (t); i >= i##end; --i) using namespace std; inline LL read() { LL x = 0, f = 1; char ch = getchar(); for(; !isdigit(ch); ch = getchar()) if(ch == ‘-‘) f = -f; for(; isdigit(ch); ch = getchar()) x = 10 * x + ch - ‘0‘; return x * f; } const int maxn = 5e5 + 10; int n, k; LL a[maxn]; int root[maxn], ch[maxn << 6][2], idx[maxn << 6], v[maxn << 6], ToT; void Insert(int &x, int pre, LL val, int LG, int id) { x = ++ToT; ch[x][0] = ch[pre][0]; ch[x][1] = ch[pre][1]; v[x] = v[pre] + 1; idx[x] = idx[pre]; if(LG == -1) { idx[x] = id; return; } int px = (val >> LG) & 1; px = px ? 1 : 0; Insert(ch[x][px], ch[pre][px], val, LG - 1, id); } int query(int x, int pre, LL val, int LG) { if(LG == -1) return idx[x]; int px = (val >> LG) & 1; px = px ? 1 : 0; if(v[ch[x][!px]] - v[ch[pre][!px]] > 0) return query(ch[x][!px], ch[pre][!px], val, LG - 1); else return query(ch[x][px], ch[pre][px], val, LG - 1); } struct Node { int l, r, mxpos, i; LL ret; Node() {} Node(int _l, int _r, int _i) { l = _l, r = _r, i = _i; mxpos = query(root[r], root[l - 1], a[i], 31); ret = a[i] ^ a[mxpos - 1]; } bool operator < (const Node &b) const { return ret < b.ret; } }; priority_queue<Node> q; int main() { // freopen("2.in","r",stdin); n = read(), k = read(); rep(i, 1, n) a[i] = a[i - 1] ^ read(); rep(i, 1, n) Insert(root[i], root[i - 1], a[i - 1], 31, i); rep(i, 1, n) q.push(Node(1, i, i)); LL ans = 0; while(k--) { Node cur = q.top(); q.pop(); ans += cur.ret; if(cur.l < cur.mxpos) q.push(Node(cur.l, cur.mxpos - 1, cur.i)); if(cur.r > cur.mxpos) q.push(Node(cur.mxpos + 1, cur.r, cur.i)); } cout << ans << endl; }
T2
给一个字符串,给 $n_a$ 个 $A$ 类子串,$n_b$ 个 $B$ 类子串,给出 $m$ 个 $A \rightarrow B$ 的连边,存在 $A_i \rightarrow A_j$ 当且仅当存在 $B$ 使得 $A_i \rightarrow B$ 且 $B$ 是 $A_j$ 的前缀
$|S|,n_a,n_b,m \leq 2 \times 10^5$
原文地址:https://www.cnblogs.com/Kong-Ruo/p/10670357.html
时间: 2024-10-09 16:54:31