可持久化可以维护数据结构的历史版本。
对于一个字典树, 如果更改一个元素, 暴力做法是复制一个树, 让后在树上修改。其实, 这样是有很多个一定一样的点是浪费的, 真正被修改的是 \(\log_2n\) 个点, \(2\log_2n\) 条边。
优点 : 大大减低时间复杂度，还支持在线做。
缺点 : 不能传懒标记, 不知道懒标记是修改哪个版本留下的
可持久化0/1 Trie, 例题 : P3835
#include<bits/stdc++.h>using namespace std;const int N = 5e5 + 5, IP = 1e9;struct TRIE{int cnt, son[2];void clear(){cnt = son[0] = son[1] = 0;}
}trie[N * 62];int tot, ans, x, p, n, v, op, c, root[N], rak;int push_back(){trie[++tot].clear();return tot;
}void Insert(int p, int q, int x){for(int i = 30; ~i; i--){for(int j = 0; j <= 1; ++j){if(!trie[q].son[j]){trie[q].son[j] = push_back();}}c = bool((1 << i) & x);trie[p].son[c] = push_back();trie[p].son[!c] = trie[q].son[!c];p = trie[p].son[c], q = trie[q].son[c];trie[p].cnt = trie[q].cnt + 1;}
}void Erase(int p, int q, int x){for(int i = 30; ~i; i--){for(int j = 0; j <= 1; ++j){if(!trie[q].son[j]){trie[q].son[j] = push_back();}}c = bool((1 << i) & x);trie[p].son[c] = push_back();trie[p].son[!c] = trie[q].son[!c];p = trie[p].son[c], q = trie[q].son[c];trie[p].cnt = trie[q].cnt - 1;}
}int S3(int p, int x){ans = 0;for(int i = 30; ~i; i--){c = bool((1 << i) & x);if(c){ans += trie[trie[p].son[0]].cnt;}p = trie[p].son[c];}return ans + 1;
}int S4(int p, int x){ans = 0;for(int i = 30; ~i; i--){if(trie[trie[p].son[0]].cnt >= x){p = trie[p].son[0];}else{x -= trie[trie[p].son[0]].cnt;p = trie[p].son[1];ans |= (1 << i);}}return ans;
}int query(int p, int q, int x){for(int i = 30; ~i; i--){c = bool((1 << i) & x);p = trie[p].son[c], q = trie[q].son[c];}return trie[q].cnt - trie[p].cnt;
}int main(){cin >> n;root[0] = push_back();for(int i = 1; i <= n; i++){cin >> v >> op >> x;if(op <= 2){root[i] = push_back();}else{root[i] = root[v];}if(op == 1){Insert(root[i], root[v], x + IP);}if(op == 2){if(query(root[i], root[v], x + IP)){Erase(root[i], root[v], x + IP);}else{root[i] = root[v];}}if(op == 3){cout << S3(root[i], x + IP) << '\n';}if(op == 4){cout << S4(root[i], x) - IP << '\n';}if(op == 5){rak = S3(root[i], x + IP) - 1;cout << S4(root[i], rak) - IP << '\n';}if(op == 6){rak = S3(root[i], x + IP + 1);cout << S4(root[i], rak) - IP << '\n';}}return 0;
}
可持久化线段树, 例题 : P3834
#include<bits/stdc++.h>using namespace std;const int N = 2e5 + 5;struct Node{int l, r, lson, rson, data;
}t[25 * N];int tot, n, m, root[N], a[N], b[N], l, r, k;int push_back(){t[++tot] = {0, 0, 0, 0, 0};return tot;
}void renew(int p){t[p].data = t[t[p].lson].data + t[t[p].rson].data;
}void build(int p, int l, int r){t[p].l = l, t[p].r = r;if(l == r){return;}int mid = (l + r) >> 1;t[p].lson = push_back(), t[p].rson = push_back();build(t[p].lson, l, mid);build(t[p].rson, mid + 1, r);renew(p);
}void updata(int p1, int p2, int ok){t[p2] = t[p1];if(t[p2].l == t[p2].r){t[p2].data++;return;}int mid = (t[p1].l + t[p1].r) >> 1;if(ok <= mid){t[p2].lson = push_back();updata(t[p1].lson, t[p2].lson, ok);}else{t[p2].rson = push_back();updata(t[p1].rson, t[p2].rson, ok);}renew(p2);
}int query(int p1, int p2, int ok){if(t[p2].l == t[p2].r){return t[p2].l;}int mid = (t[p2].l + t[p2].r) >> 1;if(t[t[p2].lson].data - t[t[p1].lson].data >= ok){return query(t[p1].lson, t[p2].lson, ok);}return query(t[p1].rson, t[p2].rson, ok - (t[t[p2].lson].data - t[t[p1].lson].data));
}int main(){ios::sync_with_stdio(0), cin.tie(0);cin >> n >> m;root[0] = push_back();build(1, 1, n);for(int i = 1; i <= n; ++i){cin >> a[i];b[i] = a[i];}sort(b + 1, b + n + 1);for(int i = 1; i <= n; ++i){a[i] = lower_bound(b + 1, b + n + 1, a[i]) - b;root[i] = push_back();updata(root[i - 1], root[i], a[i]);}for(int i = 1; i <= m; ++i){cin >> l >> r >> k;cout << b[query(root[l - 1], root[r], k)] << '\n';}return 0;
}
