标签:
http://www.lydsy.com/JudgeOnline/problem.php?id=4539
今天测试唯一会做的一道题。
按题目要求,如果暴力的把模板树往大树上仍,最后得到的大树是$O(n^2)$级别的,不能存储,更不能做了。
把模板树往大树上扔的过程我想象成了两个大节点进行连边,每个大节点代表模板树本身或一部分。
这相当于把初始的大树(此时和模板树相同)缩成一个大节点,每次把模板树的一部分缩成一个大节点往大节点构成的大树上连,最后连好的大节点构成的模板树是$O(n)$级别的。
每个节点里都套着一棵树,像树套树的模型。
这样在求距离和LCA的时候就可以先找到节点在哪个大节点里,求出在大节点内的一部分距离后,再在大节点构成的大树上倍增到LCA,统计距离。在LCA(大节点)中套着的树中继续倍增求LCA,统计距离。
每个大节点要维护的信息比较多;因为要按编号顺序从小到大加,所以还要在模板树的dfs序上用主席树查询第k大;倍增时还要注意特判几种特殊情况。在这里不再一一赘述。
测试时只有60分,后来拿到数据亲测一遍后发现到后期大树的节点到达$O(n^2)$级别,此时节点编号用int已经存不下了,所以存节点编号需要用long long。看来我还是too naive!!!
时间复杂度$O(nlogn)$
#include<cstdio> #include<cstring> #include<algorithm> using namespace std; typedef long long ll; const int N = 100003; int in() { int k = 0, fh = 1; char c = getchar(); for(; c < ‘0‘ || c > ‘9‘; c = getchar()) if (c == ‘-‘) fh = -1; for(; c >= ‘0‘ && c <= ‘9‘; c = getchar()) k = (k << 3) + (k << 1) + c - ‘0‘; return k * fh; } ll inll() { ll k = 0; int fh = 1; char c = getchar(); for(; c < ‘0‘ || c > ‘9‘; c = getchar()) if (c == ‘-‘) fh = -1; for(; c >= ‘0‘ && c <= ‘9‘; c = getchar()) k = (k << 3) + (k << 1) + c - ‘0‘; return k * fh; } int n, m, q; namespace Small { struct node { int nxt, to; } E[N << 1]; int cnt = 0, point[N], f[N][18], deep[N], L[N], R[N], a[N]; bool vis[N]; void ins(int u, int v) { E[++cnt].nxt = point[u]; E[cnt].to = v; point[u] = cnt; } void dfs(int x) { vis[x] = true; L[x] = ++cnt; a[cnt] = x; for(int i = point[x]; i; i = E[i].nxt) if (!vis[E[i].to]) { deep[E[i].to] = deep[x] + 1; f[E[i].to][0] = x; dfs(E[i].to); } R[x] = cnt; } struct node2 { int l, r, s; } T[N * 20]; int root[N], tot = 0; void update(int &pos, int l, int r, int key) { T[++tot] = T[pos]; pos = tot; ++T[pos].s; if (l == r) return; int mid = (l + r) >> 1; if (key <= mid) update(T[pos].l, l, mid, key); else update(T[pos].r, mid + 1, r, key); } void BuildPT() { for(int i = 1; i <= n; ++i) { root[i] = root[i - 1]; update(root[i], 1, n, a[i]); } } void work() { int u, v; for(int i = 1; i < n; ++i) { u = in(); v = in(); ins(u, v); ins(v, u); } cnt = 0; memset(vis, 0, sizeof(vis)); deep[1] = 0; dfs(1); for(int j = 1; j < 18; ++j) for(int i = 1; i <= n; ++i) { f[i][j] = f[f[i][j - 1]][j - 1]; } BuildPT(); } int Sum(int x) {return R[x] - L[x] + 1;} int kth(int left, int right, int l, int r, int key) { if (l == r) return l; int mid = (l + r) >> 1, sum = T[T[right].l].s - T[T[left].l].s; if (sum >= key) return kth(T[left].l, T[right].l, l, mid, key); else return kth(T[left].r, T[right].r, mid + 1, r, key - sum); } int Query(int rt, int k) { int l = L[rt], r = R[rt]; return kth(root[l - 1], root[r], 1, n, k); } int todeep(int a, int b) {return deep[a] - deep[b];} int LCA(int u, int v) { if (deep[u] < deep[v]) swap(u, v); int d = deep[u] - deep[v]; for(int i = 17; i >= 0; --i) if ((1 << i) & d) u = f[u][i]; if (u == v) return d; for(int i = 17; i >= 0; --i) if (f[u][i] != f[v][i]) { u = f[u][i]; v = f[v][i]; d += (1 << (i + 1)); } return d + 2; } } namespace Big { ll c[N][18]; int f[N][18], deep[N]; int tablenum = 0, table_to[N], table_rt[N]; ll table_l[N], table_r[N], up; int pos(ll x) { int mid, left = 1, right = tablenum; while (left < right) { mid = (left + right) >> 1; if (x < table_l[mid]) right = mid - 1; else if (x > table_r[mid]) left = mid + 1; else return mid; } return left; } void work() { int a; ll b; up = n; tablenum = 1; table_l[1] = 1; table_r[1] = n; table_rt[1] = 1; for(int i = 1; i <= m; ++i) { a = in(); b = inll(); int P = pos(b); table_rt[++tablenum] = a; table_l[tablenum] = up + 1; up += Small::Sum(a); table_r[tablenum] = up; f[tablenum][0] = P; deep[tablenum] = deep[P] + 1; table_to[tablenum] = Small::Query(table_rt[P], b - table_l[P] + 1); c[tablenum][0] = Small::todeep(table_to[tablenum], table_rt[P]) + 1; } for(int j = 1; j < 18; ++j) for(int i = 1; i <= tablenum; ++i) { f[i][j] = f[f[i][j - 1]][j - 1]; c[i][j] = c[i][j - 1] + c[f[i][j - 1]][j - 1]; } int changeu, changev, posu, vnum, posv, u, v; ll U, V, ret; for(int i = 1; i <= q; ++i) { U = inll(); V = inll(); ret = 0; posu = pos(U); posv = pos(V); if (deep[posu] < deep[posv]) { swap(posu, posv); swap(U, V); } changeu = Small::Query(table_rt[posu], U - table_l[posu] + 1); changev = Small::Query(table_rt[posv], V - table_l[posv] + 1); if (posu == posv) { printf("%d\n", Small::LCA(changeu, changev)); continue; } u = posu; v = posv; for(int i = 17; i >= 0; --i) if (deep[f[u][i]] > deep[v]) { ret += c[u][i]; u = f[u][i]; } if (f[u][0] == v) { ret += 1; u = table_to[u]; v = changev; ret += Small::LCA(u, v); ret += Small::todeep(changeu, table_rt[posu]); } else { if (deep[u] > deep[v]) { ret += c[u][0]; u = f[u][0]; } for(int i = 17; i >= 0; --i) if (f[u][i] != f[v][i]) { ret += (c[u][i] + c[v][i]); u = f[u][i]; v = f[v][i]; } ret += 2; ret += Small::LCA(table_to[u], table_to[v]); ret += Small::todeep(changeu, table_rt[posu]) + Small::todeep(changev, table_rt[posv]); } printf("%lld\n", ret); } } } int main() { n = in(); m = in(); q = in(); Small::work(); Big::work(); return 0; }
遇到看到不可做的题一定要认真地思考,看透每个操作的本质,找到操作中重复的东西并利用它化简空间复杂度和时间复杂度。一定要想好了在写,考虑到方方面面,任何细节都很关键。就像这次测试因为节点编号没有用long long存储导致$O(nlogn)$复杂度的得分被卡成$O(n^2)$复杂度的得分。
标签:
原文地址:http://www.cnblogs.com/abclzr/p/5734182.html