思路
婚姻关系建男到女的单向边,情人关系建女到男的单向边,求大小为一的强连通分量个数即可。
婚姻关系建男到女的单向边,情人关系建女到男的单向边,求大小为一的强连通分量个数即可。
这道题我们直接进行点分治即可。我们需要实现一下的几个操作:
我们一段一段来讲。
void predfs(int u, int fa) { siz[u] = 1, son[u] = 0; for (int i = head[u]; i != -1; i = edges[i].nxt) { int to = edges[i].to, w = edges[i].weight; if (to == fa || vis[to]) continue; predfs(to, u); siz[u] += siz[to], son[u] = max(siz[to], son[u]); } son[u] = max(son[u], capacity - siz[u]); if (son[u] < son[root]) root = u; }
这一段是找重心的代码,很简单我就不解释了。
void getMeta(int u, int fa, int cnt, int dist) { if (dist > k) return; ans = min(ans, sides[k - dist] + cnt); for (int i = head[u]; i != -1; i = edges[i].nxt) { int to = edges[i].to; if (to == fa || vis[to]) continue; getMeta(to, u, cnt + 1, dist + edges[i].weight); } }
这段比较重要。点分治出来之后就进行子树上的答案统计,答案的计算为\(ans=min\{sides[k – dist] + cnt\}\),其中 sides 数组代表长度为 dist 的最小段数。统计的时候不用担心边重复的问题,因为我们之后 update 操作之后才会让 sides 数组生效。
void update(int u, int fa, int cnt, int dist) { if (dist > k) return; sides[dist] = min(sides[dist], cnt); for (int i = head[u]; i != -1; i = edges[i].nxt) { int to = edges[i].to; if (to == fa || vis[to]) continue; update(to, u, cnt + 1, dist + edges[i].weight); } }
用 DFS 进行更新。
void clear(int u, int fa, int dis) { if (dis >= k) return; sides[dis] = 1e9; for (int i = head[u]; i != -1; i = edges[i].nxt) { int to = edges[i].to; if (vis[to] || fa == to) continue; clear(to, u, dis + edges[i].weight); } }
设置为极大值来覆盖子树恢复答案。
void divide(int u, int sz) { vis[u] = true, sides[0] = 0; for (int i = head[u]; i != -1; i = edges[i].nxt) { int to = edges[i].to; if (vis[to]) continue; getMeta(to, u, 1, edges[i].weight); update(to, u, 1, edges[i].weight); } clear(u, 0, 0); for (int i = head[u]; i != -1; i = edges[i].nxt) { int to = edges[i].to; if (vis[to]) continue; son[0] = capacity = (siz[to] > siz[u]) ? (sz - siz[u]) : (siz[to]); root = 0; predfs(to, 0); divide(root, capacity); } }
点分治是在重心上运行的。在重心上更新子树的信息,然后进行换根保证正确性。
// P4149.cpp #include <bits/stdc++.h> using namespace std; const int MAX_N = 200200; struct edge { int to, nxt, weight; } edges[MAX_N << 1]; int head[MAX_N], n, k, current, tmpx, tmpy, tmpz, siz[MAX_N], capacity; int ans = 1e9, sides[1002000], son[MAX_N], root; bool vis[MAX_N]; inline int read() { int re = 0, flag = 1; char ch = getchar(); while (ch > '9' || ch < '0') { if (ch == '-') flag = -1; ch = getchar(); } while (ch >= '0' && ch <= '9') re = (re << 1) + (re << 3) + ch - '0', ch = getchar(); return re * flag; } void addpath(int src, int dst, int weight) { edges[current].to = dst, edges[current].nxt = head[src]; edges[current].weight = weight, head[src] = current++; } void predfs(int u, int fa) { siz[u] = 1, son[u] = 0; for (int i = head[u]; i != -1; i = edges[i].nxt) { int to = edges[i].to, w = edges[i].weight; if (to == fa || vis[to]) continue; predfs(to, u); siz[u] += siz[to], son[u] = max(siz[to], son[u]); } son[u] = max(son[u], capacity - siz[u]); if (son[u] < son[root]) root = u; } void getMeta(int u, int fa, int cnt, int dist) { if (dist > k) return; ans = min(ans, sides[k - dist] + cnt); for (int i = head[u]; i != -1; i = edges[i].nxt) { int to = edges[i].to; if (to == fa || vis[to]) continue; getMeta(to, u, cnt + 1, dist + edges[i].weight); } } void update(int u, int fa, int cnt, int dist) { if (dist > k) return; sides[dist] = min(sides[dist], cnt); for (int i = head[u]; i != -1; i = edges[i].nxt) { int to = edges[i].to; if (to == fa || vis[to]) continue; update(to, u, cnt + 1, dist + edges[i].weight); } } void clear(int u, int fa, int dis) { if (dis >= k) return; sides[dis] = 1e9; for (int i = head[u]; i != -1; i = edges[i].nxt) { int to = edges[i].to; if (vis[to] || fa == to) continue; clear(to, u, dis + edges[i].weight); } } void divide(int u, int sz) { vis[u] = true, sides[0] = 0; for (int i = head[u]; i != -1; i = edges[i].nxt) { int to = edges[i].to; if (vis[to]) continue; getMeta(to, u, 1, edges[i].weight); update(to, u, 1, edges[i].weight); } clear(u, 0, 0); for (int i = head[u]; i != -1; i = edges[i].nxt) { int to = edges[i].to; if (vis[to]) continue; son[0] = capacity = (siz[to] > siz[u]) ? (sz - siz[u]) : (siz[to]); root = 0; predfs(to, 0); divide(root, capacity); } } int main() { memset(head, -1, sizeof(head)); scanf("%d%d", &n, &k); for (int i = 0; i < n - 1; i++) { scanf("%d%d%d", &tmpx, &tmpy, &tmpz); addpath(tmpx + 1, tmpy + 1, tmpz), addpath(tmpy + 1, tmpx + 1, tmpz); } capacity = n; son[0] = n, root = 0; for (int i = 0; i <= k; i++) sides[i] = 1e9; predfs(1, 0); divide(root, n); if (ans != 1e9) printf("%d", ans); else printf("-1"); return 0; }
首先选一个点作为根,把无根树变为有根树,然后递归进行分支(DFS-like)。但是,为了减小复杂度,我们希望在一棵树中递归层数控制到最小,那么这个点的选择就非常的重要。我们选择重心来搞定这个。有了重心,我们的子树大小都不会超过\(n/2\);计算完了之后,删掉根,并递归子树。所以时间复杂度是优秀的\(O(n\log n)\)。
重心的求法:我们用 \(O(n)\) 的时间来求出以每个点为根的子树大小,然后判断划分出联通块各点最少的点,那么这个点就是重心。
例题:IOI2011 Race,POJ1741:Tree 题解
思路:找到重心,则路径可以表示为经过根的两条子路径。第一遍 DFS 中记录走到每个点的距离和边数,同时统计到距离\(dist\)的最小边数\(sides[dist]\)。答案就是
\[ans=max\{sides[k-dist[i]]+sideSubtree[i]\}\]
一定要统计完答案之后再去用\(sideSubtree\)更新\(sides[]\)数组。
依旧需要找重心,但是在一些极限环境下(菊花图之类的)就会出锅,复杂度飙升。这个时候可以搞重构树来解决,把一颗菊花图搞成二叉树(建虚点,搞零边权的边)(目前还不怎么懂)
具体可以看这篇文章:https://ksmeow.moe/edge_based_divide_and_conquer/
例题:BZOJ2870
树链剖分是个好用的东西。我们可以第一遍 DFS 可以求出所有的重儿子,第二遍 DFS 的时候优先走重儿子,求时间戳。
对于重链的区间操作,直接暴力时间戳区间操作;对于轻边或混合链,LCA-like 算法可以搞定。
这个洛谷上题好多,不举例了。
CDQ 分治是一个神仙的离线算法。老师讲的太快,没看完就切掉了 PPT。
功能:对于所有的询问都进行二分来加速。
普通的分块最重要的一点就是确定区间长度。确定一个好的区间长度能让时间复杂度大大降低。一般我们使用\(\sqrt{n}\)来作为区间长度。用数组存好左右端点和每个点所属的区块,并预处理必要的信息。
这道题要求我们维护众数,我们可以先预处理每个两个区块间的众数,然后再每个数搞一个\(vector\)记录数出现的位置,对于暴力的区间,我们可以二分两次并相减得出出现次数。这样就可以解决问题了:
// CH4401.cpp #include <bits/stdc++.h> using namespace std; const int MAX_N = 40100; vector<int> poses[MAX_N]; int arr[MAX_N], mapping[MAX_N], lft[MAX_N], tot, rig[MAX_N], cnt[MAX_N], mode[900][900], affiliate[MAX_N]; int find(int x, int l, int r) { return upper_bound(poses[x].begin(), poses[x].end(), r) - lower_bound(poses[x].begin(), poses[x].end(), l); } void judge(int x, int l, int r, int &ans, int &ct) { int w = find(x, l, r); if (w > ct || (w == ct && x < ans)) ct = w, ans = x; } int query(int l, int r) { int ans = 0, ct = 0; int p = affiliate[l], q = affiliate[r]; if (p == q) { for (int i = l; i <= r; i++) judge(arr[i], l, r, ans, ct); return mapping[ans]; } int x = 0, y = 0; if (p + 1 <= q - 1) x = p + 1, y = q - 1; for (int i = l; i <= rig[p]; i++) judge(arr[i], l, r, ans, ct); for (int i = lft[q]; i <= r; i++) judge(arr[i], l, r, ans, ct); if (mode[x][y]) judge(mode[x][y], l, r, ans, ct); return mapping[ans]; } int main() { int n, m, t, len; scanf("%d%d", &n, &m); for (int i = 1; i <= n; i++) scanf("%d", &arr[i]), mapping[++tot] = arr[i]; sort(mapping + 1, mapping + 1 + n); tot = unique(mapping + 1, mapping + 1 + n) - (mapping + 1); for (int i = 1; i <= n; i++) arr[i] = lower_bound(mapping + 1, mapping + 1 + tot, arr[i]) - mapping, poses[arr[i]].push_back(i); t = sqrt(log(n) / log(2) * n); len = t ? n / t : n; for (int i = 1; i <= t; i++) lft[i] = (i - 1) * len + 1, rig[i] = i * len; if (rig[t] < n) lft[t + 1] = rig[t] + 1, rig[++t] = n; for (int i = 1; i <= t; i++) for (int j = lft[i]; j <= rig[i]; j++) affiliate[j] = i; memset(mode, 0, sizeof(mode)); for (int i = 1; i <= t; i++) { memset(cnt, 0, sizeof(cnt)); int ct = 0, ans = 0; for (int j = lft[i]; j <= n; j++) { if (++cnt[arr[j]] > ct || (cnt[arr[j]] == ct && arr[j] < ans)) ans = arr[j], ct = cnt[arr[j]]; mode[i][affiliate[j]] = ans; } } int last = 0; while (m--) { int l, r; scanf("%d%d", &l, &r), l = (l + last - 1) % n + 1, r = (r + last - 1) % n + 1; if (l > r) swap(l, r); last = query(l, r); printf("%d\n", last); } return 0; }
对于可以离线的问题,我们考虑询问进行分块。先对每一块内第一个问题进行暴力求解,然后块内的其他问答都用\(O(\sqrt{n})\)的时间搞定转移。非常好用。
例题:BZOJ2038: [2009国家集训队]小Z的袜子(hose)
显然,单纯计算区间内答案式子是:
\[\frac{ans}{{r-l+1 \choose 2}}, \text{其中ans是相同袜子的对数。}\]
对于每一个区块,我们可以维护一个\(num[]\)数组记录每一个颜色的出现次数,然后进行转移时对 ans 进行积累,更新询问答案即可。
// BZ2038.cpp #include <bits/stdc++.h> #define ll long long using namespace std; const ll MAX_N = 50100; struct query { ll l, r, id, answerSon, answerDom; bool operator<(const query &qy) const { return id < qy.id; } } queries[MAX_N]; ll ci[MAX_N], n, m, len, btot, num[MAX_N], lft[MAX_N], rig[MAX_N], ans; bool compare_left(query a, query b) { return a.l < b.l || (a.l == b.l && a.r < b.r); } bool compare_right(query a, query b) { return a.r < b.r || (a.r == b.r && a.l < b.l); } ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a % b); } void change(ll x, ll w) { ans -= num[x] * (num[x] - 1); num[x] += w; ans += num[x] * (num[x] - 1); } void simplify(ll id) { ll d = gcd(queries[id].answerSon, queries[id].answerDom); if (!d) queries[id].answerSon = 0, queries[id].answerDom = 1; else queries[id].answerSon /= d, queries[id].answerDom /= d; } int main() { scanf("%lld%lld", &n, &m); for (ll i = 1; i <= n; i++) scanf("%lld", &ci[i]); for (ll i = 1; i <= m; i++) scanf("%lld%lld", &queries[i].l, &queries[i].r), queries[i].id = i; len = sqrt(m), btot = m / len; sort(queries + 1, queries + 1 + m, compare_left); for (ll i = 1; i <= btot; i++) lft[i] = (i - 1) * len + 1, rig[i] = i * len; if (rig[btot] < m) lft[btot + 1] = rig[btot++] + 1, rig[btot] = m; for (ll i = 1; i <= btot; i++) { sort(queries + lft[i], queries + rig[i] + 1, compare_right); memset(num, 0, sizeof(num)); ans = 0; // previous interval; ll l = queries[lft[i]].l, r = queries[lft[i]].r; // process the first query; for (ll j = l; j <= r; j++) change(ci[j], 1); queries[lft[i]].answerSon = ans; queries[lft[i]].answerDom = (queries[lft[i]].r - queries[lft[i]].l + 1) * (queries[lft[i]].r - queries[lft[i]].l); simplify(lft[i]); for (ll j = lft[i] + 1; j <= rig[i]; j++) { while (l < queries[j].l) change(ci[l++], -1); while (l > queries[j].l) change(ci[--l], 1); while (r < queries[j].r) change(ci[++r], 1); while (r > queries[j].r) change(ci[r--], -1); if (queries[j].l == queries[j].r) queries[j].answerSon = 0, queries[j].answerDom = 1; else { queries[j].answerSon = ans, queries[j].answerDom = (r - l) * (r - l + 1); simplify(j); } } } sort(queries + 1, queries + 1 + m); for (ll i = 1; i <= m; i++) printf("%lld/%lld\n", queries[i].answerSon, queries[i].answerDom); return 0; }
把树上问题转换为序列问题,括号序列(欧拉序)可解。例题:SPOJ COT 2
给定一个长度为n的序列A[],你需要回答q个询问。每次给定两个端点,询问区间内不同颜色的个数。n, q < 1e5。
可以考虑建立主席树,以位置做版本下标,以颜色离散化后的值做下标,记录该颜色前缀个数,然后版本信息相减即可。
这道题的\(w\)很大,且要求出二维和。一种可行的办法就的是嵌套数据结构,第一想法是树状数组套线段树动态开点,时间复杂度为\(O(log^2 m)\)。但是空间依旧很大,可以发现线段树空间庞大,可以考虑替换为平衡树,空间极大减小,且时间复杂度不变。
进一步优化可以考虑对下标进行哈希。
显然需要树链剖分。
对于每一个 \(a \times dis + b\) 的操作,向懒惰标记里添加。如果碰上已经有懒惰信息,那么我们只需要把这样的操作理解为加上直线,把直线下边缘下传到左右子节点即可。(计算长度进行左右决策下传)对于每一个节点,懒惰加法为等差数列求和,计算 \(b \times subtreeSize + a \times \frac{subtreeSize*(subtreeSize-1)}{2}\)。